Every prime divisor of \(N\) must lie outside the list, since none of the \(p_i\) divides it. The always-prime option is false, as the case \(2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13+1\) shows.
The infinitude of primes is Euclid's theorem stating that the set of prime numbers is infinite: given any finite list of primes \(p_1,\dots,p_n\), there is a prime not on it.
\[p\mid p_1p_2\cdots p_n+1\ \Longrightarrow\ p\neq p_i\ \text{for all } i\]