= Solution
Put $q=p^n$. Every extension of $k=\mathbb F_p$ has <characteristic of a field>[characteristic] $p$, so the iterated <Frobenius endomorphism> satisfies
$$
(a+b)^q=a^q+b^q,
\qquad
(ab)^q=a^qb^q.
$$
The root set is $K=\{a\in F:a^q=a\}$. It contains $0$ and $1$. If $a,b\in K$, then
$$
(a-b)^q=a^q-b^q=a-b,
\qquad
(ab)^q=a^qb^q=ab,
$$
so it is closed under subtraction and multiplication. If $a\in K$ is nonzero, then
$$
(a^{-1})^q=(a^q)^{-1}=a^{-1}.
$$
Thus $K$ is closed under multiplicative inverses as well, and is therefore a <field>. This is the <polynomial characterization of a finite field>; in fact it is the <finite field> with $q$ elements, since $X^q-X$ has degree $q$ and no repeated roots.
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