Why if $n \mid m$, then $(a^n-1) \mid (a^m-1)$? My Number Theory book says that for $n, m$ be positive integers and $a>1$, then

$(a^n -1)\mid(a^m -1)$ if and only if $n\mid m$.

I understand the proof for only if part, but in if part the autor says "it is clear". However a tried to prove that but a get stuck. Can you give a hint?

So you want to show $(a^n-1) \mid (a^m-1)$ if $n \mid m$. If $n \mid m$ then $m = nk$ for some integer $k$, so you want to show $(a^n - 1) \mid (a^{nk}-1)$. 
Now use the fact that $1^k = 1$ and recall that there is a factorisation for $x^k - y^k$; one of the terms will be $a^n - 1$. 
Let me know if you need further clarification.
