The title is another homework problem I got from graduate number theory. My friend gave me this hint:
Since any two consecutive integers are relatively prime and n! and n!+1 are integers, n! and n!+1 are relatively prime. Therefore, there is at least one prime that divides n!, so there is a prime p where p < n!. Moreover, if p ≤ n, p would divide n!, which is a contradiction. Therefore, n < p ≤ n! or n+1 ≤ p ≤ n!+1.
But I am still lost. Any help would be appreciated and thanks in advance for all your time.