Why $p \leq n$ implies $p | n!$ ?
if $n > 2,$ Why $p \leq n$ implies $p | n!$ ?
I was trying to solve: prove that if $n > 2,$ then there exists a prime $p$ satisfying $n < p < n!.$
And the above statement was a part from the following hint given to the solution:
If $n! - 1$ is not a prime, then it has a prime divisor p; and $p \leq n$ implies $p|n!$.
Edit:
Why if $p | n!$ then p does not divide $n! - 1$? could anyone explain this for me please?
Hint: If $p\leq n,$ the definition of $n!$ is $n!=n\times (n-1)\times \dots \times p\times (p-1)\times \dots \times 1$ so $p|n!.$ For the big question, what is the gcd of $n!$ and $n!-1$? $(n!,n!-1)=?$ so if $p|n!-1$ can it divide $n!$?