Can someone help me prove the following:
If n is an integer and n is greater than 2, then there exists a prime p such that n is less than p which is less than n!.
Thanks so much for any help.
Now consider (n!-1), this is not divisible by any number 2<=k<=n, and so
is not divisible by any prime <=n. Hence either (n!-1) is prime or there is
some prime q, n<n<n!, which divides (n!-1). Either case is a contradiction
and so there must be a prime netween n and n!.