Prove that $b\mid a \implies (n^b-1)\mid (n^a-1)$

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.

$(n^b-1) (1 + n^b + n^{2b} + \dots + n^{(k-1)b}) = n^a - 1$

At the heart, it's the trivial identity $\rm\ 1^m \equiv 1.\ $ Since $\rm\ b\:|\:a\ $ we have $\rm\ a = b\ m\:,\ $ for some $\rm\ m\in \mathbb Z.$

Therefore $\rm \ mod\,\ n^b-1\!:\,\ n^b \equiv 1\ \Rightarrow\ n^a \equiv (n^b)^m\equiv 1^m \equiv 1,\ $ hence $\rm\ n^b-1\ |\ n^a - 1\:.$

Alternatively put $\rm\ x = n^b\ $ in $\rm\ x-1\ |\ x^m - 1,\: $ by the Factor Theorem $\rm\ x-a\ |\ f(x)-f(a)\ $ in $\rm\ \mathbb Z[x]\:.$

See this question for the special case $\rm\ x+1\ |\ x^m+1\ $ for $\rm\:m\:$ odd (follows by $\rm\ x\to -x\ $ above).

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, $(n^b-1)|(n^a-1)$.