Let $k\subset K$ be a normal extension of fields of characteristic $p>0$ with $G=\text{Aut}_k(K)$ (the automorphisms $\varphi\colon K\to K $ with $\varphi|_k = \text{id}\colon k\to k$). Show that the extension $k\subset K^G$ is purely inseparable.
So we have $k\subset K^G\subset K$ with $k\subset K$ normal, which implies that $K^G\subset K$ is normal as well. How are purely inseparable extensions and normal extensions related?
Let $a\in K^G$ have minimal polynomial $f\in k[x]$. Since $K/k$ is normal and $f$ has one root in $K$, we know that $f$ factors completely in $K[x]$, and that $G$ acts transitively on the roots of $f$. But we know that $a$ is one root of $f$ and that $\sigma(a)=a$ for any $\sigma\in G$, so therefore $a$ is the only root of $f$, or in other words, $f=(x-a)^n$ for some $n$. This is the definition of $K^G/k$ being purely inseparable.
(Note that actually, characteristic $p$ is not necessary for this argument at all. It just happens to be the case that in characteristic $0$ every extension is separable, and if $L/F$ is both separable and purely inseparable, then $L=F$.)