If $p$ is prime and $\sigma(p^k) = n$, then $p\mid (n-1)$ If $p$ is prime and $\sigma(p^k) = n$, then $p\mid (n-1)$. 
proof:  Suppose $\sigma(p^k) = [p^{k+1} -1]/(p-1) = n$.
Then $n-1 = [p^{k+1} -1]/(p-1) - 1= [p^{k+1} -1 - (p-1)] /(p-1) = [p^{k+1} - p]/(p-1) = p(p^k -1)/(p-1)$ then let $m = (p^k -1)/(p-1)$  be an integer, thus $n-1 = p\cdot m$ for some $m$. Thus, $p\mid(n-1)$.
Does this makes sense.
Please can someone please help me?
Thank you.

It is correct. $m$ is an integer because
$$ \dfrac{p^k-1}{p-1}=p^{k-1}+p^{k-2}+\ldots+p+1.$$
