I'm trying to prove that for n>2, member of Z, exists some prime p s.t. n<p<n!. I have successfully proved it by saying there's no prime btw n and (n-1)!, but I want to prove it with my original thought: first prove for 3, then for n>3: p=1+∏pi (where pi is the ith prime less ≤n) is a prime and ∏pi | ∏ni=n! and since 4 is a factor of n! but not ∏pi, ∏pi | n!/4. from there you can prove p<n! proving p>n is trickier. its easy when n factors into primes with each showing up not more than once, but I'm stuck on how to get it if n is, say, the square of two primes. I'm pretty sure p is always > n because I tried it with the first ten thousand primes with no problems.