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 $b\mid a \implies$ there exists a natural $k$ such that $bk=a$ and then raising $n$ to the power of the LHS and the RHS side 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.