I'm pondering the following number theoretical problem:
Show that if $a^m-1 \mid a^n -1$ then $m \mid n$.
My first idea was to use the factorization $a^m -1 = (a-1)(a^{m-1}+a^{m-2}+\cdots+a+1)$ in the condition $d(a^m-1)=a^n-1$, but I can't seem to bring anything out of that.
My second idea was to consider the cases where $a^m-1$ and $a^n-1$ are even/odd, using $a^m-1=2k$ or $2k+1$, respectively. But that doesn't seem to bring about anything either.
I've tried WLOG to assume $m \leq n$ and factor the divisibility condition as $da^m -d = a^n-1 \Leftrightarrow a^m(d-a^{n-m}) = d-1$, but that doesn't seem to do anything for me either.
Any ideas?