<Question>
The superfactorial $n\$$ is defined as $n\$ = \underbrace{ n!^{{n!}^{{\cdot}^{{\cdot}^{{\cdot}^{n!}}}}}}_{n!}$. What is the units digit of $4\$$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the units digit of 4\$, which is defined as a superfactorial. The superfactorial n\$ is n! raised to the power of n! , which itself is raised to n! , and so on, with the number of exponents being n! . That sounds pretty complicated, but maybe there's a pattern or a shortcut to figure out the units digit without computing the whole number.

First, let me make sure I understand the definition correctly. For n = 4, 4\$ is 4! raised to a power tower of 4! levels. Wait, no, actually, the number of exponents is n! , so for 4\$, it's 4! factorial levels? Wait, no, let me check again.

<prompt_1.1> Verify the understanding of the superfactorial definition for 4\$. </prompt_1.1>

But 4! is 24, so 4\$ is 24 24 24 ... 24 with 24 exponents? That's an enormous number. But since we only care about the units digit, maybe we can find a pattern or use modular arithmetic to simplify.

I remember that for problems involving the last digit of large exponents, Euler's theorem or Euler's totient theorem can be useful. Since we're dealing with modulus 10, the units digit is equivalent to the number modulo 10. So perhaps I can compute 4\$ mod 10.

But 4\$ is 24^24^... ^24 with 24 exponents. Let me break it down step by step.
First, let's note that 24 mod 10 is 4. So 24^anything mod 10 is the same as 4^anything mod 10. So maybe the base is 4. Then the next exponent is 24^... ^24. But even that exponent is huge. However, the exponent of 4 will determine the units digit. Since 4 cycles every 2 in its exponent pattern modulo 10.

<prompt_1.2> Analyze the units digit of 24^24^... ^24 with 24 exponents using the above plan. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The problem says n\$ is n! with n! exponents. So for 4\$, it's 4! with 4! exponents. Wait, no, maybe it's n! factorial levels? Wait, the notation is a bit confusing. Let me parse the definition again:

n\$ = n! ^{n! ^{.^{.^{n! }}}} with n! exponents. So, for example, 1\$ would be 1! = 1. 2\$ would be 2! ^{2! } = 2^2 = 4. 3\$ would be 3! ^{3! ^{3! }} = 6^{6^6}. So for 4\$, it's 4! ^{4! ^{4! ^{... ^{4! }}}}} with 4! exponents, which is 24 levels of 24. That's a power tower of 24s with 24 levels. Wait, no, 4! is 24, so the base is 24, and the exponent is another 24, and so on, 24 times. So the entire expression is 24^24^... ^24 with 24 exponents.
</execution_1.1>

<execution_1.2>
Wait, let's recall:

4^1 = 4 mod 10

4^2 = 16 = 6 mod 10

4^3 = 4*6 = 24 = 4 mod 10

4^4 = 4*4 = 16 = 6 mod 10

So the cycle is 4, 6, 4, 6, ... with period 2. Therefore, the units digit of 4^k is 4 if k is odd and 6 if k is even. So if the exponent is even, the units digit is 6; if odd, 4. Therefore, to determine the units digit of 4\$, we need to determine whether the exponent tower above 24 (which is 24^24^... ^24 with 23 more 24s) is even or odd.

But 24 is even, and any power of an even number is even. So even^even is even, and even^even^... is still even. So the exponent tower is even. Therefore, the exponent is even. Hence, 4^even = 6 mod 10. Therefore, the units digit of 4\$ is 6.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.2:
Wait, but hold on. Let me verify this step by step.

<prompt_2.1> Verify the conclusion that the units digit of 4\$ is 6 using the above plan. </prompt_2.1>

But let me check if there's a mistake here. Let's take a smaller case. Let's compute 2\$. According to the definition, 2\$ is 2! ^{2! } = 2^2 = 4. The units digit is 4. If we apply the same logic: base is 2, exponent is 2. Since 2 is even, exponent is even, so 2^even = 6 mod 10? Wait, no, 2^2 = 4, which is 4 mod 10. Wait, but according to the cycle of 2:

<prompt_2.2> Verify the method with the case of 2\$. </prompt_2.2>

Let's check 3\$. 3! is 6, so 3\$ is 6^6^6. Let's compute the units digit. 6 raised to any power is 6 mod 10. Therefore, 3\$ has units digit 6. Let's see if our method works. The base is 6, which is even, but 6^k is always 6 mod 10. So regardless of the exponent, the units digit is 6. So in this case, even if the exponent is a tower of 6s, it's still 6. So for 3\$, the units digit is 6. Similarly, 4\$ would be 24^24^... ^24 with 24 exponents. As we saw earlier, 24 mod 10 is 4, so the base is 4. Then the exponent is a tower of 24s (which is even) repeated 23 more times.

<prompt_2.3> Re-evaluate the method for 4\$ considering the base and exponent analysis. </prompt_2.3>
-----

Based on execution_1.2, execution_2.3:
But let me check 4^24. Wait, 4^24. Since 4 cycles every 2, 4^even is 6 mod 10.

<prompt_2.4> Perform a final verification of the units digit for 4\$. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Starting with 4\$, which is 24^24^... ^24 with 24 exponents. Let's denote the exponent tower as E = 24^24^... ^24 with 23 exponents. Then 4\$ = 24^E. Since 24 is congruent to 4 mod 10, we can focus on 4^E mod 10. As we saw, 4^k mod 10 is 4 if k is odd and 6 if k is even. Therefore, we need to determine if E is odd or even. But E is 24^24^... ^24 with 22 exponents. Since 24 is even, any power of 24 is even, so E is even. Therefore, 4^E mod 10 is 6. Therefore, the units digit is 6.
</execution_2.1>

<execution_2.2>
2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, so cycle length 4. Wait, so my previous thought was wrong. The cycle for 2 mod 10 is actually 4. So 2, 4, 8, 6, 2, 4, etc. So the cycle length is 4. Therefore, to compute 2^k mod 10, we need k mod 4. If k mod 4 is 1, 2, 3, 0 corresponding to 2, 4, 8, 6. But in the case of 2\$, the exponent is 2. So 2^2 = 4, which is correct. However, if we had 2^2^2, that's 2^(2^2) = 2^4 = 16, which is 6 mod 10. So here, the exponent is 4, which is 0 mod 4, so 2^4 = 6. So in that case, the exponent is 4, which is even, but the cycle length is 4, so we need to see 4 mod 4 = 0, which gives 6. So perhaps my initial approach was too simplistic.

Therefore, maybe for 4\$, even though the base is 4, which has a cycle length of 2, but the exponent is a tower of 24s. Wait, but 24 is congruent to 4 mod 10, but when considering exponents, maybe we need to look at the exponent modulo 2? Because for cycle length 2, we need to know if the exponent is even or odd. Since 4^1 = 4, 4^2 = 6, 4^3 = 4, etc. So if the exponent is even, result is 6; odd, 4. So in the case of 4\$, the exponent is 24^24^... ^24 with 23 exponents. So we need to determine if that exponent is even or odd. Since 24 is even, any power of 24 is even, so the exponent is even. Therefore, 4^even = 6 mod 10. Therefore, units digit is 6. But let's verify with a simpler case.

Take 2\$. As before, 2! = 2. Then 2^2 = 4. Correct. Now, 3\$. 3! = 6. Then 6^6^6. Since 6^anything is 6 mod 10, so 6^6^6 is 6 mod 10. So units digit 6. Similarly, 4! = 24. Then 24^24^... ^24 with 24 exponents. Since 24 is even, the base is 4, and the exponent is even. So 4^even = 6 mod 10. Therefore, units digit is 6. So perhaps the answer is 6.
</execution_2.2>

<execution_2.3>
Since the exponent is even, 4^even is 6 mod 10. So 4\$ has units digit 6. But wait, let me check if the exponent is even. The exponent is 24^24^... ^24 with 23 exponents. Since 24 is even, any power of 24 is even, so the exponent is even. Therefore, 4^even = 6 mod 10. Therefore, the answer is 6.
</execution_2.3>

<execution_2.4>
So 4^24 is 6 mod 10. Therefore, 4^(4^24) would be 4^6 mod 10. Wait, but 4^6 is (4^2)^3 = 16^3. 16 mod 10 is 6, so 6^3 mod 10 is 6. So 4^(4^24) mod 10 is 6. Then, if we have 4^(4^(4^24)) mod 10, that would be 4^6 mod 10 again, which is 6. So regardless of how many times we exponentiate, once the exponent is even, it stays 6. Therefore, even with multiple layers, if the base is 4 and the exponent is even, the result is 6 mod 10. Therefore, 4\$, which is 24^(24^... ^24) with 24 exponents, reduces to 4^(even) mod 10, which is 6. Therefore, the units digit is 6. Hmm. That seems consistent.

But to be thorough, let's check with another example. Let's take 4^4. 4^4 = 256, units digit 6. 4^4^4. The exponent is 4, which is even. So 4^(4^4) = 4^256. Since 256 is even, 4^256 mod 10 is 6. Therefore, 4^4^4 has units digit 6. Similarly, 4^4^4^4 would also be 6. So no matter how many exponents we have, as long as the base is 4 and the exponent is even, the result is 6. Since 4 is even, any power of 4 is even, so the exponent in 4\$(=24^... ) is even, hence 4^even = 6 mod 10.

Therefore, after going through this step-by-step and checking with smaller examples, I believe the units digit of 4\$/ is 6.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.3:
Wait, but let's verify with another angle. Let's consider that for modulus 10, we can use Euler's theorem. Euler's totient function φ(10) = 4, since 10 has prime factors 2 and 5, so φ(10) = 10*(1 - 1/2)*(1 - 1/5) = 10*1/2*4/5 = 4. Therefore, for any number a coprime to 10, a^4 ≡ 1 mod 10. However, 4 and 10 are not coprime, since they share a common factor of 2. Therefore, Euler's theorem doesn't apply here. So we can't reduce the exponent modulo φ(10). Instead, we need to look at the cycle length of 4 mod 10, which we saw earlier is 2. So 4^1 = 4, 4^2 = 6, 4^3 = 4, 4^4 = 6, etc.

<prompt_3.1> Re-evaluate the cycle of 4 mod 10 and its implications for 4\$. </prompt_3.1>
-----

Based on execution_2.4:
**Final Answer**
\boxed{6}
</Plan_3>

<Execution_3>
<execution_3.1>
Therefore, to compute 4^k mod 10, we only need to know if k is even or odd. If k is even, it's 6; if odd, 4. Therefore, in the case of 4\$, the exponent is 24^24^... ^24 with 23 exponents. Since 24 is even, any exponentiation of 24 will remain even, so the exponent is even. Therefore, 4^even = 6 mod 10. Therefore, the units digit is 6.

But let's check if there's a case where even exponents could lead to a different result. For example, 4^2 = 16, which is 6 mod 10. 4^4 = 256, which is also 6 mod 10. So regardless of how high the even exponent is, 4 raised to any even power mod 10 is 6. Therefore, as long as the exponent is even, the result is 6. Since the exponent in 4\$/ is 24^24^... ^24 with 23 exponents, which is even, we can confidently say the units digit is 6.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, the units digit of $4\$$ is $\boxed{6}$.
</Final_answer>
</think>