Alright my brother dug me out of my hollow to set people straight about the proof:

We assume there are finitely many primes p

_{1}...p

_{n}. We need only conclude that there exists a prime number NOT in that list, because that would disprove our original assumption that p

_{1} through p

_{n} are all the primes (it can irk you all you want but this is how a proof by contradiction works).

So consider a number q = p

_{1}*p

_{2}*...*p

_{n}+1 and consider its prime factorization (There is a theorem that states that every number factors UNIQUELY into a product of primes, or is a prime number itself). If q is a prime itself we are done, we found a prime not in our list and therefore we disprove the original assumption.

If q is not a prime it must contain a factor supposedly in our list, call it p

_{i}. Then q/p

_{i} must be an integer, but we clearly see that q/p

_{i} = (p

_{1}*p

_{2}*...*p

_{n} + 1)/p

_{i} = p

_{1}*p

_{2}*...*p

_{i-1}*p

_{i+1}*...*p

_{n} + (1/p

_{i}), and since (p

_{1}*p

_{2}*...*p

_{i-1}*p

_{i+1}*...*p

_{n}) is just a product of integers and therefore an integer itself, and (1/p

_{i}) clearly is not an integer, we conclude that p

_{i} must have been a prime not in our list in order to divide q (that is to say, q/p

_{i} cannot be an integer if p

_{i} is part of the list). Therefore q is either a prime itself or contains factors not in our list of "all the primes".

QED bitches

Hope that provides some clarification