Proof. Suppose that p1=2 < p2 = 3 < ... < pr are all of the primes. Let P = p1p2...pr+1 and let p be a prime dividing P; then p can not be any of p1, p2, ..., pr, otherwise p would divide the difference P-p1p2...pr=1, which is impossible. So this prime p is still another prime, and p1, p2, ..., pr would not be all of the primes. (Ribenboim's statement of Euclid's proof)