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I've been trying to simplify my proof of Wilson's theorem so that I can easily understand it, but I'm running into some trouble. I reached the point where I proved $(p-1)! \equiv -(p-2)! \mod p$, but I'm wondering how I can push this further. I understand that each number in $(p-2)!$ has an inverse that will cancel it out to equal 1, but is there a way of proving that $(p-2)! \equiv 1 \mod p$ in a more elementary fashion? Thanks

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    $\begingroup$ "each number in $(p−2)!$ has an inverse that will cancel it out to equal 1" is a great idea! Can you find out which numbers in $(p-2)!=1\times2\times...\times p-2$ are their own inverse? That is which values of $x$ such that $x^2=1$ in $\mathbb{Z}_p$? Once you you have found this numbers, the others cancel out as you have noticed. $\endgroup$
    – Gio
    Jan 18 at 14:48
  • $\begingroup$ Sorry, I'm just a bit confused by the wording. I thought the only inverses which are themselves are 1 and $p-1$? I cancelled these out already, but is there any other way I can prove that $(p-2)!$ is congruent to 1 mod p? I feel like the fact that inverses cancel each other out doesn't really stick with me yet, as the modulus operation was only introduced to us this year. $\endgroup$ Jan 18 at 15:00
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Alternative proof :

If you want another way of thinking about Wilson's theorem, here is an expeditive proof :

  1. Check that in $\mathbb{Z}/p\mathbb{Z}[X]$, $$X^p-X = \prod_{x \in \mathbb{Z}/p\mathbb{Z}} X-x$$ (these two polynomials are unitary, share the same degree and the same roots in the field $\mathbb{Z}/p\mathbb{Z}$)

  2. Compare the coefficients of $X$.

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Outline to prove that $(p-2)! \equiv 1 \pmod p$ (thanks to above comment from Gio and this thread):

  1. For prime $p$, every positive integer $a \in Z_p$ has a unique "multiplicative inverse" $x \in Z_p$, that is, $ax \equiv 1 \pmod p$.

So for each integer $a$ in 1 to $(p-2)$, which has its multiplicative-inverse $b$ different from itself, you can combine $a$ and $b$ and we will have $(ab) \bmod p = 1$.

  1. Now we are left with integers $a$ which are their own multiplicative-inverse. That is: $a^2 \equiv 1 \pmod p$. But that means, $p | (a-1)(a+1)$. Since $p$ is prime, this is only possible for $a=1$ and $a=p-1$. So, among 1 to $(p-2)$, there is only one such $a=1$.

(For Wilson's Theorem, we can directly use above arguments to say that: $(p-1)! \bmod p = (p-1) \bmod p = (p-1)$.)

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