Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$ Suppose $p = 2^n - 1$ is prime. Show that $n \: | \: 2^n - 2$, or equivalently $n \: | \: p - 1$.
With hint: The order of any element in this field divides $p-1$.
Example: $n=7, \; p=127 \:\Rightarrow\: 7 \: | \: 126$.

I don't have any progress that seems to lead to a proof, but here are some of my ideas:


*

*Because $p$ is a Mersenne prime, $n$ must be prime

*Use Fermat's little theorem

*Show that $2^n - 2 \equiv 0 \mod n$


You have mentioned ingredients that lead to a proof. As you point out, the number $n$ must be prime, so we can use Fermat's Theorem.
There are a couple of different versions of Fermat's Theorem. Each version is not difficult to derive from the other.
Version 1: If $n$ is prime, then for any integer $a$, we have $a^n \equiv a \pmod n$.  This says directly that $n$ divides $a^n-a$, which is exactly what you want to show, in the special case $a=2$.
Version 2: If $n$ is prime, and $n$ does not divide $a$, then $a^{n-1} \equiv 1 \pmod n$. 
In your case, $n$ is odd, so if $a=2$, then $n$ does not divide $a$. It follows that $n$ divides $2^{n-1}-1$, and therefore $n$ divides twice this number, which is $2^n-2$. 
However, you were given a hint which enables us to bypass Fermat's Theorem. So probably you were expected to argue as follows. 
Since $p=2^n-1$, certainly $p$ divides  $2^n-1$, so $2^n \equiv 1 \pmod p$.  That implies that the order of $2$ in the field $\mathbb{Z}_p$ is a divisor of $n$. But the order of $2$ is exactly $n$, since $n$ is prime, and therefore the only divisors of $n$ are $1$ and $n$. 
The order of any element of the field $\mathbb{Z}_p$ divides $p-1$. Since the order of $2$ in this field is $n$, we conclude that $n$ divides $p-1$, that is, $n$ divides $2^n-2$.  
