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An exercise in a book I'm working through asks to prove Wilson's theorem (for any prime $p$, $(p−1)!\equiv−1\pmod p$ ) using Fermat's little theorem and one of Vieta's formulas ($c_1c_2\cdots c_n=(-1)^n\frac{a_0}{a_n}$, where $c_i$ are the roots of the polynomial, and $a_0$ and $a_n$ its constant and leading coefficient, respectively). My idea was to do the following: consider the polynomial $(x-1)(x-2)\cdots (x-(p-1))$. If we were to expand it, it would look something like $x^{p-1}-\frac{p(p-1)}{2}x^{p-2}+\cdots +(p-1)!$ and if we were to consider it reduced $\pmod p$, it would hopefully become $x^{p-1}+(p-1)!$, because all of the other coefficients would ve congruent to $0$, all hopefully being multiples of $p$. Then by Fermat's little theorem, $x^{p-1}+(p-1)!\equiv 1+(p-1)!$ and if this last expression were congruent to $0$, i.e. $1+(p-1)!\equiv 0$, then $(p-1)!\equiv -1$ and we have our proof.

My question is, how do I fill in the gaps in this, i.e. that all the other coefficients are multiples of $p$ and that the final expression is congruent to $0$?

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    $\begingroup$ It does seem to be true in small cases, and it also does fail if $p$ is not a prime. Interesting. $\endgroup$ – Arnaud Mortier Dec 2 '19 at 15:44
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    $\begingroup$ It is claimed here in the introduction but apparently without a proof $\endgroup$ – Arnaud Mortier Dec 2 '19 at 15:48
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This is a standard approach. You can find details in Landau's book Number Theory (Part I, Chapter V, Theorem 77, p. 52) or perhaps other standard references.

Proof. Let $f(X) = \prod_{i=1}^{p-1} (X-i)-X^{p-1}+1$ over the field $\mathbb Z/p$. It is a polynomial of degree $p-2$, and it has $p-1$ roots, namely $1,\ldots,p-1$, so it is identically zero.

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