Question

Prove that for all integers n, if n > 2 then there is a prime number p such that n < p < n!.

Answer 1

Answer:

For all 2<k≤n it follows that k|n! so no k|(n!−1). So either n!−1 is prime or there is a prime number greater than n that divides n!−1.

Either way there is a prime number between n and n!.