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I'm trying to prove that $(p-1)! \equiv - 1 \ \text{(mod $p$)}$ for some prime $p$.

From reviewing other answers, I've seen people cite a theorem stating that $(p-1)! \equiv p - 1 \ \text{(mod $p$)}$. However, I have not before seen this theorem, nor is it cited in my textbook prior to this theorem, so I don't believe I'm able to use it. I'm therefore hoping there is some other way to prove this without using this fact.

The best I've been able to do so far is to write that $(p-1) = 1 \cdot 2 \cdot 3 \cdot \ldots \left(p-1\right)$, and since $p \equiv 0 \ \text{(mod $p$)}$, $p - 1 \equiv - 1 \ \text{(mod $p$)}$. For this to work, then, the remaining terms must either be $1$ or $-1$ with an even number of $-1$'s for the negative sign to win out. I don't believe we can apply Fermat's theorem, as each of these terms lack an exponent, even though it's clear that, since $p$ is greater than each of these terms from $1$ to $p-2$, it can't divide any of that.

Any hints or insights on how to approach this problem would be greatly appreciated.

1 Answers1

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This is actually straightforward. As you know Multiplicative group of integers modulo p i.e. $Z_p$ is an abelian group and every element has inverse, so all the pair of inverses will cancel each other. But, $p-1$ is its own inverse as $(p-1)^2 \equiv p^2 -2p +1 \equiv 1 mod(p)$

And $p-1$ can be only such element with order 2 as for any other $x \in Z_p : x^2 \equiv1 mod(p)$

$\implies p|(x+1)(x-1) \implies$ Either x=1 or x=p-1

so you finally get

$(p-1)! \equiv p-1 \equiv -1 mod (p)$

Shak
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