Let $K^{\text {perf}}$ denote the perfection of the field $K$ , which is the minimal perfect field extending $K$. Consider characteristic $p > 0$. Since it can be shown that a field $K$ of characteristic $p$ is perfect if and only if the Frobenius homomorphism $K \rightarrow K$ is an isomorphism.
$\overline K$ denote an algebraic closure of $K$, this motivates the definition
$K^{\text{perf}} := \{α \in \overline K| \exists n \in \Bbb N $ such that $\alpha^{p^n}\in K\}$:
Show that $K^{\text {perf}}$ it's a perfect field , that for every intermediate extension $K \subseteq L \subseteq K^{\text {perf}}$, $L$ is perfect if and only if $L = K^{\text {perf}}$. Compute $[K^{\text{perf}} : K]$ and the separable degree of $L/K$ denoted $[K^{\text {perf}} : K]_s$
I have already shown that is a field, but I having trouble with the rest. To prove perfectness, I have to show that every algebraic extension is separable, I'm stuck here
Could you show me how it's done?