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.

Let 
$$m=\prod_{k=1}^r p_k^{\alpha_k},\qquad p_k\ne2$$
the primary decomposition of $m$ and since $\gcd(2,p_k^{\alpha_k})=1$ then $\overline 2$ is invertible in $\Bbb Z_{p_k^{\alpha_k}}$ so there's $\mu_k\in\Bbb N$ such that
$$2^{\mu_k}\equiv 1\pmod {p_k^{\alpha_k}}$$
Now we conclude using The chinese remainder theorem and we take $n=\sum\limits_{k=1}^r \mu_k$.
