Can powers of primes be perfect numbers? 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.
 A: A quick way to prove that no prime power is perfect is to notice that, if $p$ is a prime, then $p \geq 2$, so that
$$\frac{\sigma(p^k)}{p^k} = \frac{1 + p + \ldots + p^k}{p^k} = \frac{p^{k+1} - 1}{p^k(p - 1)} < \frac{p^{k+1}}{p^k(p - 1)} = \frac{p}{p - 1}.$$
Now, since $p \geq 2$, we get that
$$\frac{1}{p} \leq \frac{1}{2} \Longrightarrow -\frac{1}{p} \geq -\frac{1}{2} \Longrightarrow 1 - \frac{1}{p} \geq 1 - \frac{1}{2} = \frac{1}{2}.$$
Consequently, we have:
$$\frac{p - 1}{p} \geq \frac{1}{2} \Longrightarrow \frac{p}{p - 1} \leq 2.$$
We conclude that:
$$\frac{\sigma(p^k)}{p^k} < \frac{p}{p - 1} \leq 2.$$
In fact, this inequality shows that all prime powers are deficient. Hence, no prime power is perfect.
A: 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.
A: An easier proof involving no inequalities may go as follows:
Let p be a prime number. Assume p^n is a perfect number where n is some natural number.
This means the sum of the factors of p^n can be written as p^n
As the top proof says, the sum of the factors of p^n can be shown as the sum of the powers of p below it all the way down to the power of zero(otherwise known as 1).
Consider the sum of the factors of p^n other than 1. One can see intuitively that it is some quantity of p, but it can also be shown with the converse of the distributive property.
That is, {p^(n-1)+p^(n-2)+...+p^2+p} can also be written as
p(p^(n-2)+p^(n-3)+...+p+1)
With this one can see this sum is divisible by p, for p is one of its factors.
Now consider the 1 we left out. If we include it in the sum, we can now say the sum of the factors of p^n are indivisible by p, for there is now a remainder of 1.
But recall the assumption that the sum of the factors of p^n can be shown to be p^n. 
With the same converse as before we can show this to be equal to 
p(p^(n-1))
This shows that it is also divisible by p, for it has p as a factor
But this is a contradiction, as it was just shown that the sum of the factors of p^n are not divisible by p, and therefore the original assumption was incorrect. A power of a prime cannot be perfect, exactly as it was meant to be shown.
