Proof by contradiction: There are infinitely many primes I need some help with a proof. I just need to be pointed in the right direction, because I've been looking at this for ages and it's not clicking. 
I need to prove that there are infinitely many prime numbers, by contradiction. The original statement is:
For all $n$ in $\mathbb{N}$ where $n > 2$, there exists a $p$ in $\mathbb{P}$[prime] such that $n < p < n!$.
We were given the hint that we're supposed to use cases to solve this. Case one is that $n!-1$ is prime, whereby obviously the statement holds. 
case two is that $n!-1$ is composite, which is somehow also supposed to prove the statement, and I don't understand how. I know that every natural number $> 1$ has at least one prime factor, but I don't understand how we know that that prime factor is greater than n. 
I also don't really understand how to do these cases using contradiction. Maybe I wrote the contradiction wrong, but I thought it came out to :
For all $p$ in $P$, there exists an $n$ in $\mathbb{N}$ where $n > 2 $, such that $n \le  p \le n!$.
But maybe I did that wrong? 
Can anyone give me a pointer on how to tie this together or where to start? I'd really appreciate it, thank you. 

Can $(2 \times 3 \times \dots \times 15 \times 16) - 1$ be divisible by 2?  No, it is clearly odd.
Can $(2 \times 3 \times \dots \times 15 \times 16) - 1$ be divisible by 3?  No, it is 1 less than a multiple of 3.
Can $(2 \times 3 \times \dots \times 15 \times 16) - 1$ be divisible by p, when $p \le 16$?  No, it is 1 less than a multiple of p.
So the first prime it $16! - 1$ can be divisible by must be at least 17.
