Euler's totient function Proof that $n\mid \varphi(2^n-1)$

Question

Let $\varphi$ denote Euler's totient function, i.e., for $k\in\mathbb N$, $\varphi(k)$ is the number of natural numbers less tan $k$ which are relatively prime to $k$.

Prove that for every $n\in\mathbb N$ we have $n\mid \varphi(2^n-1)$.
Hint: Compute $o([2])$ in $U_{2^n-1}$, the group of units of $\mathbb Z_{2^n-1}$.

I am totally stuck here for this question. how do I compute o([2])??

any tips or suggestions would be great!

Answer 1

Note that trivially $$2^n\equiv 1\mod 2^{n}-1$$

Now suppose $k<n$. Then $2^k-1<2^n-1$, so...?