<Question>
For how many positive integers $n>1$ is it true that $2^{24}$ is a perfect $n^{\text{th}}$ power?
</Question>

<think>
<Plan_1>
Okay, so I need to figure out for how many positive integers n > 1 the number 2^24 is a perfect nth power. Hmm, let me start by understanding what the question is asking. 

A perfect nth power means that there exists some integer k such that k^n = 2^24. So, essentially, 2^24 must be equal to k^n. My task is to find all n > 1 where this equation holds true for some integer k.

<prompt_1.1> Understand the problem statement and set up the equation for k^n = 2^24. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me write that down:

k^n = 2^24

I need to find all integers n > 1 such that this equation is satisfied for some integer k.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
First, maybe I should recall some exponent rules. If k^n = 2^24, then k must be a power of 2 as well, right? Because 2 is a prime number, so its only factors are powers of itself. So, let me assume that k is 2^m for some integer m. Then substituting back into the equation:

<prompt_2.1> Assume k is 2^m and substitute back into the equation k^n = 2^24. </prompt_2.1>
-----

Alternatively, another way to think about this is that the exponents must divide each other when you take roots. For 2^24 to be a perfect nth power, the exponent 24 must be divisible by n, right?

<prompt_2.2> Consider an alternative method to verify the solution by checking if 24 is divisible by n. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
(2^m)^n = 2^24

Simplifying the left side, when you raise a power to a power, you multiply the exponents:

2^(m*n) = 2^24

Since the bases are the same and the equation holds, the exponents must be equal. Therefore:

m*n = 24

So, m and n are positive integers such that their product is 24. But n has to be greater than 1, as per the problem statement.
</execution_2.1>

<execution_2.2>
Wait, no, that might not be exactly correct. Let me think. If 2^24 is a perfect nth power, then 24 must be a multiple of n. Because when you take the nth root of 2^24, you get 2^(24/n). For that to be an integer, 24/n must be an integer. Therefore, n must be a divisor of 24. 

Ah, so that's another way to look at it. So n must be a divisor of 24, and n > 1. Therefore, the number of such n is equal to the number of divisors of 24 minus 1 (since we exclude n=1).
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, hold on. Let me check that again. If m*n = 24, then n must be a divisor of 24. Because m has to be an integer, so n divides 24. Therefore, the possible values of n are exactly the positive divisors of 24, excluding 1 since n > 1.

Therefore, the number of such n is equal to the number of divisors of 24 minus 1.

But let me verify this.

<prompt_3.1> Verify the initial reasoning by listing the divisors of 24 and checking if they satisfy the condition. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's list the divisors of 24. The prime factorization of 24 is 2^3 * 3^1. Therefore, the number of divisors is (3+1)*(1+1) = 4*2=8. So there are 8 divisors. Let's list them: 1, 2, 3, 4, 6, 8, 12, 24. Excluding 1, we have 7 divisors. Therefore, there are 7 possible values of n.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me make sure I didn't skip any steps here. Let's go back.

<prompt_4.1> Re-evaluate the steps to ensure no details were skipped and confirm the solution. </prompt_4.1>
-----

Based on execution_2.2:
Let's confirm that.

<prompt_4.2> Verify the alternative method by listing divisors of 24 and checking if they satisfy the condition. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
If k is 2^m, then n must divide 24. So n can be any divisor of 24 greater than 1. Therefore, the number of such n is equal to the number of divisors of 24 minus 1 (to exclude n=1). Since 24 has 8 divisors, the answer is 7. That seems correct.

But let me check with an example. Take n=2. Then 2^24 is (2^12)^2, so yes, that's a perfect square. Similarly, n=3: 2^24 is (2^8)^3, since 8*3=24. That works. n=4: (2^6)^4=2^24. Yep. n=6: (2^4)^6=2^24. n=8: (2^3)^8=2^24. n=12: (2^2)^12=2^24. n=24: (2^1)^24=2^24. All of these are valid, and n=2,3,4,6,8,12,24, which are 7 numbers. So that checks out.
</execution_4.1>

<execution_4.2>
First, find all divisors of 24. As before, 1, 2, 3, 4, 6, 8, 12, 24. Then exclude 1, so 7 divisors. Therefore, the answer is 7. 

But wait, hold on. Let me check if both approaches are equivalent. The first approach considered m*n=24, so n must divide 24. The second approach says that n must divide 24. So yes, they are the same. Therefore, both methods lead to the same conclusion.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But let me think again: is there any other possible n >1 where 2^24 is a perfect nth power but n does not divide 24?

<prompt_5.1> Consider if there are any other possible values of n where 2^24 is a perfect nth power. </prompt_5.1>
-----

Based on execution_4.2:
Therefore, the answer is 7. 

But just to be thorough, let me check each divisor greater than 1:

<prompt_5.2> Perform a final verification of the solution by checking each divisor greater than 1. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
Let's see. Suppose there exists an n that does not divide 24 but 2^24 is still a perfect nth power. For example, suppose n=5. Then 2^24 would have to be a perfect 5th power. That would require 24 to be divisible by 5, which it isn't. Similarly, n=7: 24 isn't divisible by 7. So, no. Therefore, n must divide 24. Hence, the only possible n are the divisors of 24 greater than 1.
</execution_5.1>

<execution_5.2>
For n=2: 24/2=12, which is an integer. So 2^12 is the 2nd root of 2^24.

n=3: 24/3=8, integer. 2^8 is the 3rd root.

n=4: 24/4=6, integer. 2^6.

n=6:24/6=4, integer. 2^4.

n=8:24/8=3, integer. 2^3.

n=12:24/12=2, integer. 2^2.

n=24:24/24=1, integer. 2^1=2.

All of these are valid integers. Therefore, all these n satisfy the condition. 

Therefore, the number of positive integers n>1 is 7.
</execution_5.2>
</Execution_5>

<Final_answer>
Therefore, there are $\boxed{7}$ positive integers $n > 1$ for which $2^{24}$ is a perfect $n^{\text{th}}$ power.
</Final_answer>
</think>