TITLE: $n > 2$ implies $n < p < n!$ - Proof
QUESTION [3 upvotes]: So, I am having trouble approaching this question: 

Prove that for all natural numbers $n > 2$, there exists a prime number $p$ such that $n < p < n!$ 

I have been thinking about Bertrand's postulate for some time; however, that theorem cannot be taken as a given in this question. So, I have to actually prove Bertrand's postulate first, before continuing towards the actual claim in the question. 
Note: Direct proof, proof by induction, contraposition, exhaustion (cases), contradiction, etc. all ideas and contributions are welcome.

REPLY [5 votes]: Let $n$ be a natural number greater than 2. Let $p$ be any prime number
that divides $n!−1$. Since $p\mid n!−1$, $p$ does not divide $n!$. It follows
that $p$ is not any natural number less than or equal to $n$, and so $p$ is a natural
number greater than $n$. Also, $p$ is less than or equal to $n!−1$, since $p$ divides
$n!−1$. It follows that $n < p < n!$, so there is a prime between $n$ and $n!$.

REPLY [1 votes]: You can use a variant of Euclid's proof of the infinitude of the primes:
Suppose $p_1, p_2, \dots, p_k$ are the primes that are less than or equal to $n$. Then, consider $p_1 p_2 \cdots p_k + 1$.
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