elementary number theory - If $a$ divides $b$, then $(2^a-1)$ divides $(2^b-1)$

Prove that if $a \mid b$, then $(2^a-1) \mid(2^b-1)$.

So, I've said the following:

1. Let $a \mid b$.


2. $\implies b=am$


3. Assume $(2^a-1) \mid (2^b-1)$


4. $\implies (2^b-1)=(2^a-1)x$

But I can't figure out what to do with any of this from here forward.