Every element of a finite field of order q satisfies x^q = x

Statement

Let $F$ be a finite field of order $q$. Then $a^q = a$ for every $a \in F$, and $$x^q - x = \prod_{a \in F} (x - a) \quad \text{in } F[x].$$

Remarks

The proof: the group of units $F^\times$ has order $q - 1$, so $a^{q-1} = 1$ for $a \neq 0$, by Lagrange's theorem. Then $a^q = a$ for all $a$, also for $a = 0$. So the $q$ elements of $F$ are $q$ distinct roots of $x^q - x$, which has degree $q$. This gives the factorisation. For $q = p$ prime, $a^p = a$ in $\mathbb{Z}/p\mathbb{Z}$ is Fermat's little theorem.

Explanations

L1 human

In a field with $q$ elements, raising to the power $q$ changes nothing. So the polynomial $x^q - x$ has every element of the field as a root.