Prove $n\mid \phi(2^n-1)$ If $2^p-1$ is a prime, (thus $p$ is a prime, too) then $p\mid 2^p-2=\phi(2^p-1).$
But I find $n\mid \phi(2^n-1)$ is always hold, no matter what $n$ is. Such as $4\mid \phi(2^4-1)=8.$
If we denote $a_n=\dfrac{\phi(2^n-1)}{n}$, then $a_n$ is A011260, but how to prove it is always integer?
Thanks in advance!

$(\mathbb{Z}/\mathbb{(a^n-1)Z})^*$ is a group of order $\phi(a^n-1)$ and gcd$(a,a^n-1)=1\Rightarrow a\in (\mathbb{Z}/\mathbb{(a^n-1)Z})^*$ 
We have  $a^n\equiv 1\mod (a^n-1)$ and $a^k-1<a^n-1$ when ever $k<n$ so there does not exist $k<n$ such that the above condition holds. So the order of $a$ in $(\mathbb{Z}/\mathbb{(a^n-1)Z})^*$ is $n$. And as the order of each element divides the order of the group so we have $n|\phi(a^n-1)$
Putting $a=2$ we have the required result asked in the question.
