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 $0$?