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!

I will use Lifting the Exponent Lemma(LTE).
Let $v_p(n)$ denote the highest exponent of $p$ in $n$.
Take some odd prime divisor of $n$, and call it $p$.
Let $j$ be the order of $2$ modulo $p$.
So, $v_p(2^n-1)=v_p(2^j-1)+v_p(n/j)>v_p(n)$ as $j\le p-1$.
All the rest is easy. Indeed, let's pose $n=2^jm$ where $m$ is odd.
Then $\varphi\left(2^{2^jm}-1\right)=\varphi(2^m-1)\varphi(2^m+1)\varphi(2^{2m}+1)\cdots\varphi\left(2^{2^{j-1}}m+1\right)$. At least $2^j$ terms in the right side are even.
