Show a prime number exists between $n$ and $n!$

Question

All $n$ bigger than $2$ has $p$ (prime number) such that $p$ divides $n$. Then a $p$ exists which divides $(n - 1)!$

With this, i have no idea how to show that $p$ is bigger than $n$ ($n < p < n!$).

Answer 1

$(n-1)!$ doesn't work because its prime factors are all less than $n$.

Instead, look at $n!-1$ which is coprime to any numbers $k=2,3,\dots,n$. Thus a prime factor of $n!-1$ is larger than $n$ but less than $n!$.

Answer 2

Use Bertrand Postulate hence $(2n-2)<n!$ you get your proof