rabbit, you just proved that there can't be FOUR primes. We're proving that there can't be a FINITE number of primes. You can't say that for p1...pn; how can you say pn !| 211 when you don't know what pn is?
Anyways, here's the proof. Figure it's about time to say it. (Even though I said Sunday.)
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This will be proven by contradiction.
Assume that there is a finite number of primes, p1p2p3...pn, which comprise set P. Consider Q := p1p2p3...pn + 1. If Q were composite, then it would have to be divisible by some product of primes in P. This does not mean, however, that the left-hand (L | L= p1p2p3...pn) and right-hand (R | R = 1) terms have to be divisible by some prime in P. It does mean, however, that L / x + R / x must be an integer. Since L / x will always be an integer, as it contains all the primes in our set P, and R / x will never be an integer, since 1 is only divisible by 1, which is not a member of P, we will be adding an integer + a non-integer. Therefore Q is not divisible by x, and therefore Q is prime.
QED
I lost.
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I never actually saw an "official" solution to this problem, so feel free to criticize my proof!
Edit: corrected some semantics. The proportion thing didn't achieve what I wanted it to, but the integer + non-integer thing does.