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I am not sure how to start this queston. Could you give me some hint?

Show that for any prime number $p$, every prime divisor of $p!+1$ is greater than $p$.

Thanks

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    $\begingroup$ Some terrorist decided to downvote everything here. Did what I could to heal the damage. $\endgroup$ – punctured dusk May 16 '17 at 14:32
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Why don't you start by assuming that there is a prime divisor which is less than or equal to $p$ and try to show contradiction?

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Notice that all primes less than or equal to $p$ divide $p!$. Therefore, they cannot also divide $p!+1$. Therefore, $p!+1$ can be represented as a product of primes which are all greater than $p$. This is actually one of the nice proofs showing that there are infinitely many primes.

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the key idea here is that any two consecutive integers are coprimes: for any positive integer $n$ we have: $(n,n+1)=1$.

Thus, $p!+1$ and $p!$ have no common divisor (apart from $1$). But all primes less than $p$ are (by the definition of the factorial) divisors of $p!$ thus none of them can be a divisor of $p!+1$.

Thus: the least prime divisor of $p!+1$ must necessarily be greater than $p$.

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Here's a hint: If $q$ is any prime (in fact, any positive number) less than or equal to $p,$ and you divide $p!+1$ by $q,$ what is the remainder?

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Hint. If $q\leq p$ then $q\mid p!$

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