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How might one show that $p$ divides $(p-2)!-1$ where $p$ is a prime number? I am not even sure if it is true but I have been randomly trying it on some primes and it seems to be true.

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    Do you know Wilson's Theorem? $(p-1)!\equiv -1\pmod{p}$. Hence $(p-2)!(p-1) \equiv (p-1)\pmod{p}$, and cancelling $p-1$ (which is relatively prime to $p$) we get $(p-2)!\equiv 1\pmod{p}$.2012-01-25
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    Did you mean "$p$ divides $(p-2)!-1$"?2012-01-25
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    @MichaelHardy: of course you are right, careless me. :)2012-01-25
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    @ArturoMagidin: Interesting theorem, thanks!2012-01-25
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    Does not compute. Only 1 and p dividies p for p prime.2012-01-25
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    @all: [Laylady's comment] answers an earlier version of the question (which asked why $(p-2)!-1$ divides $p$).2012-01-25

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