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How would I go about proving this? I do not understand the $ p | n! $ and $ p | n! -1 $ approach ... which I have found on previous posts.

dasboat
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  • This was asked almost 3 years ago: https://math.stackexchange.com/questions/1084296/for-all-n2-there-exists-a-prime-number-between-n-and-n – projectilemotion Oct 29 '17 at 21:17

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Since $n!=1\cdot 2\cdot 3\cdot ...\cdot n$, we know that every number from $2$ to $n$ divides $n!$.

This means that if we subtract $1$ from $n!$, then the result is not divisible by any number from $2$ to $n$.

So, any prime factor of $n!-1$ has to be larger than $n$ ... but of course smaller than $n!$.

Bram28
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  • Right! This is a (perhaps simpler) variation of Euclid's proof that there are infinitely many primes. – lhf Oct 24 '17 at 00:41