= Solution
Write $q=p^r$, where $p$ is the <characteristic of a field>[characteristic]. In a <splitting field> of $X^{q^m}-X$ over $F$, let $K$ be its set of roots. The <derivative> of this <polynomial> is $-1$, so all $q^m$ roots are distinct. The <Frobenius endomorphism>[Frobenius] <power function of a statistical test>[power] is additive and multiplicative: $(x+y)^{q^m}=x^{q^m}+y^{q^m}$ and $(xy)^{q^m}=x^{q^m}y^{q^m}$. Thus roots are closed under addition, negatives and multiplication; for a nonzero root, $(x^{-1})^{q^m}=x^{-1}$. They include $0,1$ and form a <field>.
Every $a\in F$ satisfies $a^q=a$, by <Lagrange's theorem for finite groups>[Lagrange's theorem] on $F^\times$ and separately at zero. Iterating gives $a^{q^m}=a$, so $F\subseteq K$. The constructed <field> therefore contains $F$ and has exactly $\boxed{q^m\text{ elements}}$.
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