If $p$ is prime and $\sigma(p^k) = n$, then $p\mid (n-1)$

Question

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.

Answer 1

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