Hey bkbowser.
The proof is very good. Very simple and very elegant.
The latex outputs look goofy in my browser. I'm not sure whats up with that so I'm abandoning it.
Suppose that p=2^k-1 is prime. Prove that either k=2 or k is odd.
Proof;
Assume p is prime and n,p,k are integers.
p=2^k-1
assume k=2 then p=3 which is prime. this concludes the first case.
assume k is even. k=2n then,
p=2^{2n}-1=(2^n+1)(2^n-1)
Since k=2n>2, n>1 and both of the factors of p, (2^n+1) and (2^n-1) are greater then 1 for all n
Since p is a prime it has two factors p and 1
since neither factor of p can be 1, p can not be a prime if k is even.
This is a contradiction with the premise that p is a prime, by reductio ad absurdum we conclude that it is not the case that k is even.
Since k is a natural number it can only be even or odd. And as k is not even it must be odd.