Solution
= Solution
By the allowed finite multiplicative-<subgroup> theorem, the <group> $K^\times$ is cyclic. Let $z$ generate it. Every nonzero element of $K$ is a power of $z$, hence belongs to $F(z)$; zero belongs as well. Since the reverse inclusion is automatic,
$$
\boxed{K=F(z).}
$$
Thus a finite extension of a <finite field> is a <simple field extension>.