Okay, so I need to find the units digit of 4$$, which is a superfactorial. The superfactorial is defined as n$$ = n!^(n!^...^n!) with n! number of n!'s. That seems complicated, but maybe I can break it down step by step.

First, let me make sure I understand the definition correctly. For n$$, we take n factorial, then raise it to the power of n factorial, and continue this n! times. So for 4$$, it's 4!^(4!^(4!^...^4!)) with 4! number of 4!'s. Let me compute 4! first. 4! is 4×3×2×1 = 24. So 4$$ is 24^(24^(24^24)) with 24 exponents. Wait, that's a tower of 24s, all 24s, 24 levels high. That's an astronomically large number. But I don't need the actual number, just the units digit.

The key here is that when dealing with units digits, especially in exponents, there's a pattern that repeats every certain number. For example, the units digit of powers of 2 cycles every 4: 2, 4, 8, 6, 2, 4, etc. Similarly, for other digits, there's a cycle length. So maybe I can find the units digit of 24^something modulo 10, but since 24 is the base, let's see. Wait, 24 modulo 10 is 4, right? So 24 ≡ 4 mod 10. Therefore, 24^k ≡ 4^k mod 10. So the units digit of 24^k is the same as the units digit of 4^k.

Therefore, if I can find 4^k mod 10, where k is the exponent in the tower. But k itself is 24^(24^(24^24)) with 23 more 24s. Wait, no. Wait, 4$$ is 24^(24^(24^24)) with 24 exponents. Wait, 4! is 24, so 4$$ is a tower of 24s, 24 layers high. So the exponent is 24^(24^(24^...^24)) with 23 more 24s. So the exponent is 24^(24^(24^...)) with 24 total exponents. Wait, maybe I need to clarify.

Wait, the definition says n$$ is n! raised to itself n! times. So for n=4, n!=24, so 4$$ is 24^24^24^...^24 with 24 exponents. So it's a power tower of 24s, 24 levels high. So the exponent k is 24^(24^(24^(...^24))) with 23 more 24s. So k = 24^(24^(24^...^24)) with 23 exponents. That's a huge exponent. But since we're dealing with mod 10, maybe we can find k mod something.

But first, since we have 4^k mod 10, and 4^k has a cycle of 2. Let's check: 4^1=4, 4^2=16 (units digit 6), 4^3=64 (units digit 4), 4^4=256 (units digit 6), so the cycle is 4,6,4,6,... every 2 exponents. So the units digit of 4^k is 4 if k is odd, and 6 if k is even. So we need to determine whether k is odd or even.

But k is 24^(24^(...^24)) with 23 exponents. Let's see. Let's figure out k's parity. Since 24 is even, any power of 24 is even. So 24^anything is even. Therefore, 24^(24^(...)) is even. So k is even. Therefore, 4^k has units digit 6. Therefore, 24^k ≡ 4^k ≡ 6 mod 10. So the units digit is 6. Wait, is that right?

Wait, hold on. Let me verify. Let's take it step by step.

First, 4$$ = 24^(24^(24^...^24)) with 24 exponents. So the exponent in the tower is 24^(24^(...^24)) with 23 exponents. Let's denote this exponent as E. So E = 24^(24^(...^24)) with 23 exponents. Then 4$$ = 24^E. But since 24 ≡ 4 mod 10, 24^E ≡ 4^E mod 10. So we need to find 4^E mod 10.

The units digit of 4^E is 4 if E is odd, 6 if E is even. So we need to determine if E is odd or even. E is 24^(24^(...^24)) with 23 exponents. Since 24 is even, 24 raised to any power is even. Therefore, E = 24^(something) is even. Therefore, E is even. Therefore, 4^E ≡ 6 mod 10. Therefore, the units digit is 6. So is the answer 6?

Wait, but maybe there's a step I missed here. Let's check with smaller n. For example, take n=2. Then 2$$ = 2!^(2!) = 2^2 = 4. Units digit 4. Correct. For n=3, 3$$ = 6^(6^6). Let's compute the units digit. 6^anything is 6, so 6^(6^6) has units digit 6. Then 6^6 has units digit 6. So 6^(6^6) = 6^6 mod 10 = 6. So the units digit is 6. Wait, but 6^(anything) is 6, so regardless of the exponent, 6^k ≡ 6 mod 10. So for n=3, the units digit is 6.

Similarly, for n=4, 4$$ = 24^(24^(24^24)). So as before, 24^E. But 24 ≡ 4 mod 10, so 4^E. Since E is even, 4^E ≡ 6 mod 10. So units digit 6. So same as n=3. Wait, but n=2 gives 4. Hmm. So maybe for n >=3, the units digit is 6. Let's check n=1. 1$$ = 1!^(1!^...^1!) with 1! exponents. Since 1! is 1, so 1^1 =1. Units digit 1. So n=1:1, n=2:4, n=3:6, n=4:6. So maybe starting from n=3, the units digit is 6. So for n=4, it's 6. So is that the answer?

But let's verify with n=3. 3$$ =6^(6^6). Let's compute the exponent 6^6. 6^6 is 46656. Then 6^46656. But as I said, 6^k is always 6 mod 10. So units digit is 6. Similarly, for n=4, 24^E, where E is even. So 4^E mod 10 is 6. So units digit is 6. So seems like for n >=3, the units digit is 6. So answer is 6.

But let me check again. Wait, 4$$ is 24^E, where E is 24^(24^(...^24)) with 23 exponents. So E is 24^something. Since 24 is even, 24^anything is even. So E is even. Therefore, 4^E mod 10 is 6. Therefore, the units digit is 6.

But maybe I should check with a different approach. Let's compute the exponent modulo 4, since the cycle of 4^k mod 10 has period 2. So if we can find E mod 2, then we can determine if E is even or odd. Since 4^E mod 10 is 4 if E is odd, 6 if E is even.

But E is 24^(24^(...^24)) with 23 exponents. Let's compute E mod 2. Since 24 is even, 24 ≡ 0 mod 2. So 24^anything ≡ 0 mod 2. Therefore, E = 24^(something) ≡ 0 mod 2. Therefore, E is even. Therefore, 4^E ≡ 6 mod 10. So yes, units digit is 6.

Alternatively, since E is even, regardless of how large E is, 4^E will end with 6. So 24^E ends with 6. Therefore, the answer is 6.

Wait, but let me think again. Suppose n=3. 3$$ =6^(6^6). The exponent is 6^6. Let me compute 6^6 mod something? Wait, 6^k mod 10 is always 6, so 6^6 mod 10 is 6. Therefore, 6^6 ends with 6. Then 6^(6^6) ends with 6. So yes, same as before.

But maybe for n=4, the exponent is 24^(24^(...^24)) with 23 exponents. Let's see. Let's compute E mod 4. Wait, because 4^k mod 10 cycles every 2, but maybe we need to compute modulo something else?

Wait, no. Since we are dealing with 4^E mod 10, and 4^k cycles every 2. So to find 4^E mod 10, we need to know E mod 2. Since 4^E mod 10 is 4 if E is odd, 6 if E is even. So E's parity is all we need. As before, E is 24^(24^(...^24)) with 23 exponents. Since 24 is even, 24^anything is even. Therefore, E is even. Therefore, 4^E ≡6 mod 10. So units digit is 6.

Therefore, the answer is 6. So I think that's the conclusion.

But let me check with n=5 as a test case. 5$$ =120^(120^(120^(120^120))) with 120 exponents. Then 120 ≡0 mod 10. So 0^k mod 10 is 0 if k>0. But wait, 0^1=0, 0^2=0, etc. So 0^anything is 0. Therefore, 120^anything is 0 mod 10. But wait, 120^E is 0 mod 10, but 120 is 0 mod 10. So 120^E is 0 mod 10. So units digit 0. But according to our previous logic, n=5 would have units digit 0. But according to our earlier logic, n>=3, units digit is 6. Wait, contradiction here.

Wait, so maybe my earlier reasoning is flawed. Let's check.

Wait, for n=5, 5! =120. So 5$$=120^(120^(120^...^120)) with 120 exponents. So the base is 120, which is 0 mod 10. Then 120^anything is 0 mod 10, except when the exponent is 0. But since the exponent is a tower of 120s, which is 120^(...), which is positive, so the exponent is at least 120. Therefore, 120^120 is 0 mod 10. Therefore, 5$$ is 0 mod 10. Therefore, units digit is 0.

But according to our previous logic, n>=3, the units digit is 6. But here it's 0. So maybe the rule is different. Wait, so when the base is 0 mod 10, the result is 0 mod 10, regardless of the exponent (as long as exponent is positive). Therefore, if n! ≡0 mod 10, then n$$ ≡0 mod 10. So when is n! ≡0 mod 10?

n! is divisible by 10 when n >=5, since 5! =120. So for n >=5, n! is divisible by 10, so n!^k ≡0 mod 10 for any k >=1. Therefore, n$$ = (n!)^(n!^...) with n! exponents. Since n! >=120, so the exponent is >=120. Therefore, n!^(something) ≡0 mod 10. Therefore, units digit is 0.

But for n=4, 4! =24 ≡4 mod 10. So 24^k ≡4^k mod 10. Then, the units digit depends on k's parity. For n=4, as we saw, the exponent is even, so 4^k ≡6 mod 10. Therefore, units digit is 6.

But for n>=5, n! is divisible by 10, so n$$ ≡0 mod 10. So units digit is 0.

Therefore, the answer for 4$$ is 6, and for n>=5, it's 0. Therefore, the answer is 6.

Therefore, to confirm, 4! =24. The exponent is a tower of 24s, 24 levels high. Since 24 is even, the exponent is even. Therefore, 4^even ≡6 mod 10. So units digit is 6. Correct.

But wait, 24^something where something is even. 24 ≡4 mod 10. So 4^even ≡6 mod 10. So yes, units digit 6. So answer is 6.

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