Prove $\forall n > 2, \ \exists p\in \Bbb{P} : n < p < n!$ I need to prove that:

$$(1) \ \forall n\in\Bbb{N}_{\ge2}, \ \exists p\in \Bbb{P} : n < p < n!$$

I already know how to prove the $n < p$ part; it directly follows from the proof that there is no largest prime.  However, I am stumped on the $p < n!$ part.
One idea I had for showing this is as follows: we know that $(2) \ \forall n \in \Bbb{N}_{\ge2},\ \exists m\in\Bbb{N} : n!=2m$. By testing a few values, I conjectured that $(3) \ \forall n\in\Bbb{N}_{\ge2}, \ \exists p\in \Bbb{P} : n < p < 2n$. If (3) could be shown, it would be simple to prove (1), but there does not seem to be no easy way to prove (3), if it's even correct.
 A: You can obviously proceed by stating the Bertrand's postulate. However, I think the question is asking you to solve it a bit differently.
Let us solve it using a method similar to that used by Euclid when he tried to prove that there are infinite number of primes.
Let $p_{1},p_{2},\cdots ,p_{k}$ be all the primes less than or equal to $n$. Obviously, $k<n$.Let, $P=p_{1}\times p_{2}\times\cdots \times p_{k}+1$.
Notice that $P$ is not divisible by any of the given primes. Hence $P$ is either a prime or is divisible by a prime $> n$.
It can be easily seen that $$P<n!$$ Furthermore, $P$ needs to be greater than $n$, otherwise we would find another prime (other than $p_{1},p_{2},\cdots,p_{k}$) which would lead to a contradiction.
A: What you speculate on is correct, there is always a prime between $n$ and $2n$.  This is known as Bertrand's postulate and is really a theorem.  But for between $n$ and $n!$ you can modify the Euclid proof for a much easier approach.  It is true for $3, 3 \lt 5 \lt 6$.  For $n \gt 3,$ take the product of all primes less than $n$ and add $1$.  This is less than $n!$ because it has only one factor of $2$, not the three coming from $2$ and $4$.  Then it is either prime or composite ...
A: Just consider $n! - 1$.  Clearly this is between $n$ and $n!$ as long as there is anything between $n$ and $n!$, and it isn't divisible by any prime $p \le n$.  Thus it contains a prime factor bigger than $n$ and smaller than $n!$.
