Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 11F (Number Theory)

Reduce the polynomial modulo $p$, obtaining a nonzero degree-$d$ polynomial over the field ${\mathbb{F}}_p$. If $a$ is a root, polynomial division gives $\bar{f}(X)=(X-a)h(X)$. Any different root $b$ is a root of $h$, because $b-a\ne 0$ is invertible. Induction on degree therefore proves there are at most $d$ distinct roots modulo $p$.

Every nonzero residue is a root of $X^{p-1}-1$ by Fermat's little theorem. The difference between this monic polynomial and the product over all its $p-1$ nonzero roots has degree at most $p-2$, yet has $p-1$ roots. The root bound forces it to vanish identically over ${\mathbb{F}}_p$. Thus all coefficients of the corresponding integer polynomial are divisible by $p$.

For the requested failure modulo $p^2$, take $p=7$: $6!+1=721=7\cdot 103$, and $103$ is not divisible by seven. Hence the prime-square strengthening fails, even though some special primes do satisfy it.

Compare constant terms in the polynomial identity proved above. The constant term of $X^{p-1}-1$ is $-1$, while that of the product is $(-1{)}^{p-1}(p-1)!$. For odd $p$, this gives $(p-1)!\equiv -1(\mathrm{mod}p)$. For $p=2$, the same conclusion holds because $1\equiv -1(\mathrm{mod}2)$. Thus Wilson's theorem follows:

For odd $p$, pair the terms with denominators $j$ and $p-j$. Since both denominators are coprime to $p$,has numerator divisible by $p$ and denominator not divisible by $p$. Summing these $(p-1)/2$ pairs over a common denominator coprime to $p$ gives a numerator divisible by $p$. Cancelling common factors cannot remove a factor $p$, since none occurs in the denominator. Therefore the numerator of $u_p$ in lowest terms is divisible by $p$. Equivalently, inversion permutes the nonzero residues and ${\sum }_{j=1}^{p-1}{j}^{-1}\equiv {\sum }_{j=1}^{p-1}j\equiv 0(\mathrm{mod}p)$.