## Every finite field has prime power order

Let $F$ be a finite field of characteristic $p$. Prove that $|F| = p^n$ for some $n$.

Suppose $F$ has a nonzero element $\alpha$ of additive order $q$. That is, $q$ is minimal such that $q\alpha = 0$. Note that $q \leq p$, since $p$ annihilates every element of $F$ (being the characteristic of $F$). If $q < p$, then by the division algorithm in $\mathbb{Z}$ we have $p = qb + r$ for some $b$ and $r$ with $0 \leq r < q$. If $r \neq 0$, then $r \alpha = 0$, violating the minimalness of $q$. If $r = 0$, then $p = qb$. Since $p$ is prime, either $q = 1$ (so $\alpha = 0$) or $b = 1$ (so $q = p$), which both give a contradiction. Thus $q = p$.

So every element of $F$ has additive order $p$, and thus by Cauchy’s theorem $|F|$ is a power of $p$.