Can powers of primes be perfect numbers?

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.

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.