Now \(N\) is certainly larger than \(p\text{.}\) Also, \(N\) is not divisible by any number less than or equal to \(p\text{,}\) since every number less than or equal to \(p\) divides \(p!\text{.}\) Thus the prime factorization of \(N\) contains prime numbers (possibly just \(N\) itself) all greater than \(p\text{.}\) So \(p\) is not the largest prime, a contradiction. Therefore there are infinitely many primes.
in-context