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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.

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    `case two is that n!-1 is composite` Then it has a prime factor. Can that prime factor be $\le n\,$? – 2017-02-25
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    @dxiv Oh wow. That should be an answer. – 2017-02-25
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    @S.C.B. Hi , I noticed that in the negation you edited "n>=p>=n!" to "n<=p<=n!" can you tell me why you just added the "=" to each "<" instead of flipping them? I thought negating an inequality flipped the signs.. thanks so much for your help – 2017-02-25
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    @BabaSvoloch The fact that it was a negation was pretty unclear. Are you trying to derive a contradiction, as $n!>n$ for most $n$? – 2017-02-25
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    @S.C.B. All we were told is 1) to do a proof by contradiction, and 2) to use the cases i. n!-1 is prime, and ii. n!-1 is composite. I'm not really sure what I'm trying to do from there. If I was unclear it's a reflection of the fact that I'm struggling to understand how to do it. I apologize for the confusion – 2017-02-25

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Case 1, as you already said, is trivial.

Case 2 is a little more tricky. Suppose that n!-1 is composite. Then it must be divisible by at least two primes, as you have already stated. But since n! is divisible by all numbers less than n, consider- what numbers less than n could go into n!-1?

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    I assume that what you're getting at is that no numbers less than n can go into n!-1. I've written out a few examples but I still don't understand why. – 2017-02-25
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    @BabaSvoloch All the numbers up to $n$ divide $n!$, so they can't also divide $n! {-} 1$ – 2017-02-25
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    @BabaSvoloch Yes, that's exactly where I would take the problem. If a prime p less than n goes into n!, it can't also divide n! - 1, but since all primes less than n divide n!... There aren't many options left over for our dear friend n! - 1. – 2017-02-26
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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.