Prove that for all odd $n$, there is an $m$ such that $2^m - 1$ is divisible by $n$ I've been trying to solve a problem that reads as such

Prove that for all odd positive integers $n$, there exists a positive integer $m$ such that $(2^m) - 1$ is divisible by $n$.

Proof by induction doesn't seem like a good option, because for each n, m can be arbitrarily different.
 A: Consider the sequence $a_k=2^k$ mod $n$ since there are only finitely many residue classes mod $n$, there must be two terms which are equal
$$
2^{k_1}\equiv2^{k_2}\pmod{n}\tag{1}
$$
since $n$ is odd, $\frac{n+1}2\times2\equiv1\pmod{n}$, that is $\frac{n+1}2$ is the multiplicative inverse of $2$. Therefore, we can multiply both sides of $(1)$ by $\left(\frac{n+1}2\right)^{k_1}$ to get
$$
1\equiv2^{k_2-k_1}\pmod{n}\tag{2}
$$
which means that
$$
\left.n\,\middle|\,2^{k_2-k_1}-1\right.\tag{3}
$$
A: How much theory are you allowed to use? $2$ is invertible in $\mathbb{Z} / n \mathbb{Z}$, as $n$ is odd, so $2$ is in the group of invertible elements of $\mathbb{Z} / n \mathbb{Z}$, and you may take $m$ to be its period in this group.
A: One considers that set of numbers $2^a \mod{n}$.  
Now these leave various remainders, which we shall assume to be different.  Since $2^a$ is an integer not a multiple of $n$, the number of remainders must be less than $n$.
Now suppose that $2^{a+1}$ is already in the abvoe set, say $2^b$.  Now, $n \mid 2^{a+1} - 2^b$. Since $m$ is odd, then $n \mid 2^{a+1-b}-2^{b-b}$, whereapon, $m=a+1-b < n$, and if $b$ were the first instance of a repeat of remainders, then $b=0$ and $2^{a+1}=1$.  
A: If this problem comes from a number theory textbook or course, I'm assuming they want you to use Fermat-Euler Theorem. 
n is odd so: $$\gcd(2,n) = 1
\implies 2^{\varphi(n)} \equiv 1 \pmod{n}\implies 2^{\varphi(n)}-1 \equiv 0 \pmod{n}$$
