TITLE: Prove that $b\mid a \implies (n^b-1)\mid (n^a-1)$
QUESTION [11 upvotes]: Given natural numbers $a,b,n$, prove $b\mid a \implies (n^b-1)\mid (n^a-1)$.

I tried the simple method of beginning with $b\mid a \implies \exists k \in \mathbb{N} $ such that $bk=a$ and then raising $n$ to the power of the LHS and the RHS and eventually forming $(n^b)^k-1=n^a-1$. That's obviously not enough. It tried making it work from the other side and didn't get too far either. I guess there's some $gcd$ theorem or something I need.
Any ideas?
Thanks.

REPLY [3 votes]: There is a very useful identity 
$$
x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-1}).
$$
Substituting $y$ with 1 here, we get
$$
n^a-1=(n-1)(n^{a-1}+n^{a-2}+\cdots+1).
$$
Now since $b|a$, there exists some $k\in\mathbb{Z}$ such that $a=bk$. Therefore, 
$$
n^a-1={(n^b)}^k-1=(n^b-1)（(n^b)^{k-1}+(n^b)^{k-2}+\cdots+1).
$$
Hence, $(n^b-1)|(n^a-1)$.