Prime number $x$ and $x!$

Question

I'm trying to show that there exists a prime $p$ between $x$ and $x!$ where $x \in \mathbb{N}, x>2$ such that, $$ x < p < x!$$ Can I say that since $x!$ and $x!-1$ are relatively prime(share no common divisors)there exists a prime $p>x$ s.t $p|x!-1$

Answer 1

That looks like a decent proof to me, assuming $x!>x$. If this is the entire exercise, however, I would maybe prefer it if each step was expanded a bit (at least if this was on a test or an assignment that I was correcting). For instance, we see using the Euclidean algorithm (or something to that effect) that $x!$ and $x! - 1$ are coprime. And a prime $p$ that doesn't divide $x!$ must be larger than $x$, and a prime that divides $x!-1$ must be smaller than $x!$. So since $x!-1$ has prime factors, there must be a prime between $x$ and $x!$ (this is the part of the proof that would fail for $x = 2$: $2! - 1$ doesn't have any prime factors).