Not necessarily, it might be composite, but in this case one of it's prime factors will necessarily not lie in the supposed list of primes, therefore also a contradiction.
The first counter example to "If L := {P0,P1,..,Pn} is a list of primes, then prod(L)+1 is prime" is {2,3,5,7,11,13}, their product is 30030, and 30031 is a composite of 2 primes, none of which are in the list.
For instance, 2 * 3 * 5 * 7 * 11 * 13 + 1 = 30031 = 59 * 509.
Assume p1, ..., pn is a finite list of primes. The sum p1+...+pn+1 is divisible by a prime, because every natural number> 1 is. However, it's not divisible by p1,...,pn, hence there must be an additional prime not in the list.
(I think you're right though that GP's "contradiction" doesn't work)
Never thought of using "by definition, all numbers can be divided by a prime" tu merge the two cases. It's not that shorter, but is IMHO quite elegant, I'll remember it. Thanks for correcting me.
Euclid's original proof of the theorem is of the form "for any list of primes, I can find an additional prime" [0], and for good reason: in Ancient Greece, thinking of infinity, or infinite sets, as a concrete object that you could manipulate would have seemed weird.
But the proof variant where you produce a contradiction doesn't really get into the set-theoretic details either. All it does is say: "Assume there is a finite list of all primes. Derive a contradiction. Therefore there is no such list." That's pretty much equivalent to the direct proof, it's just using different logical inference rules.
[0]: http://aleph0.clarku.edu/~djoyce/java/elements/bookIX/propIX...