Problem: Factoring $x^p-x= (x)(x-1)\cdots(x-(p-1)) \in \Bbb Z_p[x]$.
Clearly $0$ is a root, and so consider $(x^{p-1}-1) \mod p$, by Fermat's little theorem, for each $a \neq$ 0 , $(x-a)$ is a root, and each root must have multiplicity $1$ since $(x^{p-1}-1)$ has at most $p-1$ roots ($Z_p[x]$ is an integral domain).
At each step above, by the factor theorem for commutative rings: $(x^{p-1}-1)=q(x)(x-k)$ for $1 \leq k \leq p-1$ and $q(x)\in\Bbb Z_{p}[x]$, but I am not sure how to conclude that overall $(x^{p-1}-1)= (x-1)\cdots(x-(p-1))$
Hint:
Show that the $(x-a),\; a\in\mathbf Z_p$, are coprime, so that $x^p-x$ will be divisible by their product. Finally, compare the degrees and the leading coefficients.