# For all $n>2$: there exists $p$ prime: $n<p<n!$

The question is:

For all $n>2$, where $n \in \mathbb Z$: there exists $p$ prime such that $n<p<n!$

Here is my Proof:

$\forall$ $p<n: p|n!$, or $p$ divides $n!$

Since $n!$ and $n!-1$ are relatively prime

$=>$ $n!$ and $n!-1$ share no common divisors

$=>$ there must a prime $p > n$ such that $p | (n!-1)$

I feel like it needs to be clarified more. What else should I add? Thanks in advance!

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the idea is correct, you can explain why $p|n!$ for $p\leq n$ for better clarity. – Denis Sep 4 '13 at 10:18

To me, it looks correct, and I wouldn't add much more. You may want to add that $n!-1 > 2$, just in case.

Of course, the level of precision needed in an argument depends on who you are talking to and in what circumstances: a research paper would surely take things such as this problem for granted, while in a homework assignment this level of precision should be appropriate.

A much stronger statement is true. It is called Bertrard's postulate, and says that for any $n \geq 2$ you can find a prime $p$ such that $n < p < 2n$.

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Another possible proof:

Let $S=\{ x \in \mathbf{Z^+} : n<x<n! \text{ and for every prime divisor } d \text{ of } x ,d \leq n \}$ . Since $(n!-1) \in S$ , the well-ordering principle implies that exists a minimal element in $S$ , called $n_0$ . This $n_0$ is prime.

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