Factoring x^p - x over Z/pZ (Mod p Coefficients)

Let's factor the polynomial $x^p-x$ over modulo $p$ (formally $\mathbb{Z}/p\mathbb{Z}[x]$ or $\mathbb{F}_p[x]$).

So let's find some roots of the polynomial $x^p-x$. If $0$ a root? Of course: $0^p - 0 = 0 \pmod{p}$. What about $1$? Yes, too. $1^p - 1 = 0 \pmod{p}$. Hmmm, what if every number from $0$ to $p-1$ was a root? Wouldn't that be cool? Well, this is exactly what happens! But why?

If you remember Fermat's little theorem, it says that for any positive integer $a$ and prime $p$, we have $a^p \equiv a \pmod{p}$, which is equivalent to $a^p-a \equiv 0 \pmod{p}$. So actually $0, 1, 2, \ldots, p-1$ are all roots! That's $p$ roots. And the degree of $x^p-x$ is exactly $p$. So by the factor theorem, and the fact that a polynomial has at most as many roots as its degree, it factors as exactly the product of $(x-\text{roots})$.

As an equation, $$x^p-x \equiv x(x-1)(x-2)\ldots(x-(p-1)) \pmod{p}.$$ Here's an example for $p = 3$: $$x(x-1)(x-2) = x^3-3x^2+2x \equiv x^3-x \pmod{3}.$$ (You can check it for larger primes too.)