I need to prove the following, though I'm not 100% certain I understand the definition of a perfect number.
Prove that no perfect number is a power of a prime.
First of all, I'm assuming that the question is asking me to prove that for any prime $p$ and for all natural numbers\positive integers $n$, $p^n$ is not a perfect number. Am I correct in this understanding of the problem?
Based on this, I've come up with the following to prove this theorem...
Let $p$ be a prime number. Assume that $p^n$ is a perfect number for some $n\in\mathbb{N}$. Therefore, $$p^n=\underbrace{p\cdot p\cdot p\cdots p}_{n\text{ prime numbers}}=\underbrace{1+p+p^2+p^3+\cdots +p^{n-1}}_{\text{sum of all divisors of $p^n$ except itself}}=\frac{p^n-1}{p-1}.$$
As $\frac{p^n-1}{p-1}\leq p^n-1<p^n$ for all $p\geq 2$, this is a contradiction, thus proving that no power of a prime can be a perfect number.
Without elaborating too much, I'm assuming that my proof ends here, because the definition I was given for a perfect number is that it is equal to the sum of all of its divisors except itself. Since the only valid divisors of a number of the form $p^n$ are 1 and all powers of $p$ from $1$ to $n-1$, this is what I come up with. And since $1$ is not a prime number by convention, this seems to hold.
Note: I used the identity $1+x+x^2+x^3+\cdots +x^{n-1}=\frac{x^n-1}{x-1}$ because it was conveniently proven in my textbook.
As André Nicolas your reasoning is good an it's enough to prove the that no perfect number is a prime of power, but you should prove that $p^n \neq \frac{p^n - 1}{p-1}$ to complete the answer, because it's not so obivous and something we take for granted. Here's some help.
Try to prove using contradiction. Assume that:
$p^n = \frac{p^n - 1}{p-1}$
$p-1$ is obviously not 0, so we multiply by it.
$$p^n(p-1) = p^n - 1$$ $$p^{n+1} - p^n - p^n = -1$$ $$p^{n+1} - 2p^n = -1$$ $$p^n(p-2) = -1$$
Because both terms are integers, that means that we have two separate cases:
$$ \left\{\begin{aligned} &p^n = 1\\ &p-2 = -1 \end{aligned} \right.$$
This implies that $p=1$, but because p is a prime, it can't be 1.
$$ \left\{\begin{aligned} &p^n = -1\\ &p-2 = 1 \end{aligned} \right.$$
The second equation implies that $p=3$, but $3^n = -1$ isn't possible in any case. So because we exhausted all the possibilites and we didn't find a solution, that means that our initial assumption is wrong.
Q.E.D.