= Solution
Write $q=p^r$, where $p=\operatorname{char}K$. For
$$
h(X)=X^{q^n}-X,
$$
the derivative is
$$
h'(X)=q^nX^{q^n-1}-1=-1,
$$
so $h$ is a <separable polynomial> and all its roots in its <splitting field> $L$ are distinct.
Let $R$ be the set of these roots. The <Frobenius endomorphism> and the identities
$$
(x+y)^{q^n}=x^{q^n}+y^{q^n},
\qquad
(xy)^{q^n}=x^{q^n}y^{q^n}
$$
show that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne0$, inversion. Thus $R$ is a <finite field>. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ and
$$
q^{[L:K]}=|L|=q^n.
$$
Hence the <polynomial characterization of a finite field> gives
$$
\boxed{[L:K]=n}.
$$
Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an <algebraic closure>. The <minimal polynomial of an algebraic element> gives
$$
[K(\alpha):K]=n,
$$
so $K(\alpha)$ has $q^n$ elements. Its nonzero elements form a <finite group> under multiplication, of order $q^n-1$; hence $\alpha^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the <irreducible factors of a finite-field Frobenius polynomial> satisfy
$$
\boxed{f(X)\mid X^{q^n}-X}.
$$
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