, 2019

We first show that $n$ is a power of $p$. Let $n = p^k t$ where $k \ge 0$ and $t \ge 1$ are integers and $p \nmid t$. Let $M = m^{p^k}$. By the assumption in the problem $M^{p t}-1 = (M^t-1)q$ for some prime number $q$. Recall that $(M^a-1, M^b-1) = M^{(a,b)} - 1$ for all positive integers $a$ and $b$, and therefore we get $(M^p-1, M^t-1) = M-1$. Since both $M^p-1$ and $M^t-1$ divide $M^{pt}-1$ we see that $\frac{(M^p-1)(M^t-1)}{M-1}$ divides $M^{pt}-1$.
In other words, $\frac{M^p-1}{M-1}$ divides $\frac{M^{pt}-1}{M^t-1} = q$. Then, $\frac{M^p-1}{M-1}$ is either $1$ or $q$. Clearly it is not $1$, and hence it is $q$ so we get $t=1$. Now since $p$ is an odd prime number, the statement $p^{k+1}|(p-1)^{p^k}+1$ readily follows from the binomial expansion of $(p-1)^{p^k}$.