Q:

Proving any odd number is a factor of $2^n -1$ for some $n$

I'm struggling with a proof of the following. I feel like it should be a one-liner or something simple but I'm just not grasping the idea:
Suppose that m is an odd natural number. Prove that there is a natural number $n$ such that $m$ divides $2^n -1$
Any help would be much appreciated, thanks.

A:

A start: Consider the remainders when $2$, $2^2$, $2^3$, $2^4$, and so on are divided by $m$.  
By the Pigeonhole Principle there exist distinct positive integers $i\lt j$ such that $2^i$ and $2^j$ have the same remainder on division by $m$.
Show that $m$ divides $2^{j-i}-1$.

