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:(adsbygoogle = window.adsbygoogle || []).push({});

first prove for 3, then for n>3:

p=1+∏p_{i}(where p_{i}is the ith prime less ≤n) is a prime and ∏p_{i}| ∏n_{i}=n! and since 4 is a factor of n! but not ∏p_{i}, ∏p_{i}| 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.

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# Proof that exists prime btw n<p<n!

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