If $N$ is the algebraic closure of a finite field $F$, prove that $\operatorname{Gal}(N/F)$ is an abelian group and that any element of the Galois group has infinite order.
If $N$ were a finite extension of $F$, then it would be easy to see that $\operatorname{Gal}(N/F)$ is a cyclic group. But I don't know if I can proceed using this idea in the above case.
Let $F$ be your finite field, say of order $q$ (where $q$ is necessarily a power of a prime $p$). It is known from the elementary theory of finite fields that any extension $E/F$ of degree $n$ is Galois, with cyclic Galois group isomorphic to $(\Bbb Z/n\Bbb Z, +)$, generated by the so called Frobenius automorphism, sending $x$ to $x^q$. "The" algebraic closure of $F$ being the inductive limit of the finite extensions of $F$, its Galois group over $F$ is the projective limit of all the $\Bbb Z/n\Bbb Z$, usually denoted by $\widehat{\Bbb Z}$, the "profinite completion" of $\Bbb Z$. By construction (check on projective limits, or just argue that $\Bbb Z$ is dense in $\widehat{\Bbb Z}$ with respect to the profinite topology), $\widehat{\Bbb Z}$ is abelian and has no non null element of finite order.
