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.

You want to find $n$ such that
$$2^n\equiv1\mod m$$
It's clear that the values of $2^n\text{ mod } n$ start repeating after a while. Can it happen that $1$ is excluded from the cycle? No because $2$ has an inverse $\text{mod }m$, namely $(m+1)/2$, so you can go in reverse and must reach $1$ eventually.
