Galois group of $\overline{F}/F$ 
Let $K$ be the algebraic closure of $F$ where $F$ is a finite field. Show that $Gal(K/F) \simeq \hat{\mathbb{Z}}$.

I know that $\hat{\mathbb{Z}} = \varprojlim \mathbb{Z}_{n}$, so its enough to show that $Gal(K/F) \simeq \varprojlim \mathbb{Z}_{n}$. I can solve the problem if $|F|=p$, but and if $|F| = p^{n}$? Can someone help me?

More generally, Gal$(\bar F/F)$ is the inverse limit of its finite quotients, so it suffices to pick up, if possible, a convenient inductive system of finite extensions of $F$. If $F$ is a finite field (its characteristic is irrelevant), we know that every finite extension of $F$ is cyclic , with Galois group generated by the Frobenius automorphism. Hence Gal$(\bar F/F)\cong \hat {\mathbf Z}$. Of course this doesn't work all the time : we don't know the absolute Galois group of $\mathbf Q$ !
