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.
For one direction we have
Hint: Note that $$ a^n-1\mid a^{nk}-1 $$
For the other direction, suppose that $$ a^n-1\mid a^m-1\tag{1} $$ Let $m=qn+r$ where $0\le r\lt n$. We know that $$ a^n-1\mid a^{qn}-1\tag{2} $$ Subtract $(2)$ from $(1)$ to get $$ a^n-1\mid a^m-a^{qn}\tag{3} $$ Since $\gcd(a^n-1,a)=1$, $(3)$ implies $$ a^n-1\mid a^r-1\tag{4} $$ Since $0\le r\lt n$, $(4)$ implies that $r=0$ (otherwise, a larger positive integer divides a smaller one). Since $r=0$, $m=qn$; that is, $n\mid m$.
If $n|m$, say $m = dn$, then $$ a^m-1 = {a^n}^d - 1 = (a^n - 1)((a^n)^{d-1} + (a^n)^{d-2} + \ldots + a^n + 1) $$ can be checked directly or follows from setting $\alpha = a^n$ and $N = d$ in $$ \alpha^N-1 = (\alpha-1)(\alpha^{N-1} + \alpha^{N-2} + \ldots + \alpha + 1). $$
Alternatively, you could also just note that all the complex roots $\zeta$ of $x^n - 1$ must be roots of $x^m - 1$ if $n\mid m$, since $\zeta^n = 1 \implies \zeta^m = (\zeta^n)^d = 1$.
