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: Write $2^a-1$ and $2^n-1$ in binary notation: the first is a string of $a$ $1$’s, and the second is a string of $n$ $1$’s. Now think about doing an ordinary long division, still in base two. Can you see why there will be no remainder?
