= Solution
Write $|F|=p^n$, where $n=[F:\mathbb F_p]$ since $F$ is a finite-dimensional <vector space> over its <prime field>. Every nonzero element satisfies $x^{p^n-1}=1$ by <Lagrange theorem>, so $F$ is exactly the <splitting field> of $X^{p^n}-X$ over $\mathbb F_p$. The <derivative> is $-1$, making it a <separable polynomial>. Thus the extension is a <normal field extension> and a <separable field extension>, hence a <Galois extension>.
The <finite-field Frobenius automorphism> $\sigma(x)=x^p$ fixes $\mathbb F_p$ and has $\sigma^n=1$. If $\sigma^d=1$ with $0<d<n$, all $p^n$ elements would be roots of the degree-$p^d$ <polynomial> $X^{p^d}-X$, impossible. It has order $n$, equal to the degree of the <Galois extension>. Therefore
$$
\boxed{\operatorname{Gal}(F/\mathbb F_p)=\langle\sigma\rangle\cong C_n.}
$$
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