Proof: If $n=ab$ then $2^a-1 \mid 2^n-1$ I don't know how to explain or how to prove the following statement
If $n=ab$ and $a,b \in \mathbb{N}$ then $2^a-1 \mid 2^n-1$.
Any ideas? Perhaps an induction?
Thanks in advance.

Hint:  Recall that $$x^u-1=(x-1)(x^{u-1}+x^{u-2}+\cdots+x+1).$$  Then $$2^{ab}-1 =(2^a)^b-1$$ and letting $x=2^a$, $u=b$ we see that
$$(2^a)^b-1=(2^a-1)\left((2^a)^{b-1}+(2^a)^{b-2}+\cdots+(2^a)+1\right).$$  This means that $2^a-1$ must divide it.  We can use a similar argument for $2^b-1$.
