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!
 A: Hint: Clearly $2$ has order $n$, modulo $2^n-1$.

Further Hint: We know that $2^{ \phi(2^n -1)} \equiv 1 \pmod{2^n-1}$.

A: Observation. If $2^n-1\mid 2^k-1$ then $n \mid k$.
Proof. Let $2^n-1\mid 2^k-1$. Let $k=qn+r$ where $0\le r<n$. Then we get 
$$2^k-1=2^{qn+r}-1=2^{qn+r}-2^r+2^r-1=2^r(2^{qn}-1)+2^r-1.$$
Since $2^{qn}-1=(2^n-1)(2^{(q-1)n}+2^{(q-2)n}+\dots+2^n+1)$, we see that $2^n-1$ divides $2^{qn}-1$. Therefore $2^n-1$ also divides
$$2^r-1 = 2^k-1 - 2^r(2^{qn}-1).$$
But $2^n-1 \mid 2^r-1$ with $0\le r<n$ is only possible if $r=0$. So we get that $k=qn$ and $n\mid k$. 

From Euler's theorem we have
$$2^{\varphi(2^n-1)}\equiv 1 \pmod{2^n-1},$$
i.e., $2^n-1 \mid 2^{\varphi(2^n-1)}-1$. (Notice that $\gcd(2,2^n-1)=1$, so Euler's theorem can be applied here.) 
Using the above observation for $k=\varphi(2^n-1)$ we get
$$n\mid \varphi(2^n-1).$$
Note: The same argument would work to show that $n\mid \varphi(a^n-1)$ for any $a\ge2$. Some posts about this more general question:


*

*$n\mid \phi(a^{n}-1)$ for any $a>n$?

*$n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer.

*Prove that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra
A: 
Consider $U(2^n-1)$. Clearly $2\in U(2^n-1)$. It can also be shown
  easily that the order of $2$ in the group $U(2^n-1)$ is $n$. By
  Lagrange's theorem $|2|=n$  divides $|U(2^n-1)|=\phi(2^n-1)$.

Remark: Thank you for letting me know about this fact. It's interesting!
A: 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.
A: $(\mathbb{Z}/\mathbb{(a^n-1)Z})^*$ 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.
