Mersenne prime

Bleys

I was trying to prove the statement "If (2^n)-1 is prime then n is prime". I've already seen the proof using factorisation of the difference of integers and getting a contradiction, but I was trying to use groups instead. I was wondering if it's possible, since I keep getting stuck.
So far I've got:
If (2^n)-1=p where p is prime then [tex]2^{n}\equiv1 mod(p)[/tex]. The group {1,2,...,p-1} is cyclic and every element has order p-1. So n = k(p-1) for some positive integer k. But doesn't this mean n is not prime? which is wrong I know. Could someone point me in the right direction or a theorem that might be useful?

fantispug

Interesting problem... I'm not yet sure of a proof, but your statement

is wrong. All you know is that every element has an order that divides p-1. (E.g. for p=7 the element 2 has order 3).