If $m\mid n$ then $p^m-1\mid p^n-1$

Question

I know $m$ ,$n$ are two positive integer numbers such that $m\mid n$. If $p$ is a prime number, I want to show $p^m-1\mid p^n-1$.

Answer 1

Rewrite $n=d\cdot m$ and observe that we are looking to show

$p^m-1|(p^m)^d-1$.

This boils down to showing that $x-1|x^d-1$, which should not be a problem.

$\textbf{Note}$: We don't need $p$ to be prime.