 Solution

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.