number theory proof Does this proof work?
Prove or disprove that if $\sigma(n)$ is a prime number, n must be a power of a prime.
Since $\sigma(n)$ is prime, $n$ can not be prime unless it is the only even prime, $2$, since $\sigma(n)$ for prime $n=p+1$ which will always be even and therefore not a power of a prime.
Now, assume that n is a power of a prime.
Then $\sigma(n)=\sigma(p^a)$ for prime number $p$ 
so since $\sigma$ is a multiplicative function,
$\sigma(n)=\sigma(p^a)=\sigma(p)\sigma(p)...\sigma(p)$
which is not prime since it has multiple factors.
$n$ must therefore not be a power of a prime

But $\sigma(3^2)=1+3+9=13$ is prime, so it’s certainly possible that $n$ is a power of a prime when $\sigma(n)$ is prime. You have the right ingredients, but you’ve not put them together quite right.
Suppose that $\sigma(n)$ is a prime. Let the prime factorization of $n$ be
$$n=p_1^{r_1}p_2^{r_2}\ldots p_m^{r_m}\;;$$
then
$$\sigma(n)=\prod_{k=1}^m\sigma(p_k^{r_k})=\prod_{k=1}^m\frac{p_k^{r_k+1}-1}{p_k-1}\;.\tag{1}$$
Now prove that each factor in the product in $(2)$ is greater than $1$ and conclude that there can be only one factor. It immediately follows that $n=p_1^{r_1}$ is a power of a prime.
