Power of Frobenius automorphism: Why $\phi(\alpha)^n = \alpha^{p^n}$ where $\phi$ is the Frobenius automorphism I cannot prove this. Even more, I have to conclude that $K = \{\alpha \in F | \phi^n(\alpha) = \alpha$} is a subfield of $F$.
What I have trying for the second part is:
$K$ is the splitting field of $p(x) = x^{p^n} - x$. So, $K \cong GF(p^n)$ that is a subfield of $F$. Is this correct?

By definition, the Frobenius automorphism $\phi$ on some finite field $F$ of characteristic $p$ is defined to be $\phi(a) = a^p$. So by definition,
$$
\phi^2(a) = \phi(\phi(a)) = \phi(a^p) = (a^{p})^{p} = a^{p^2}.
$$
Similarly, 
$$
\phi^n(a) = \phi(\phi^{n-1}(a)) = \phi(a^{p^{n-1}}) = (a^{p^{n-1}})^p = a^{p^n}.
$$
Now the second part follows from that fact that $\phi$ and so also $\phi^n$  is a field automorphism of $F$. If $a,b \in K$, then $\phi^n(a+b) = \phi^n(a)+\phi^n(b) = a+b,$ so $K$ is closed under addition. We can similarly show that it is closed under multiplication. Multiplicative inverses will exist in $K$ since $\phi^{n}(a^{-1}) = \phi^{n}(a)^{-1} = a^{-1}.$
