Let $L/K$ be a purely inseparable extension of degree $p$. Prove that $L=K(\alpha)$ for some $\alpha \in L.$
Since $L/K$ is purely inseparable and $[L:K]=p$, the minmal polynomial of every element $a\in K$ over $L$ must be of the form $x^{p}-b$, but then I can't make any headway. That led to my second idea: let $M$ be the Galois closure of $L/K$ and $G=\operatorname{Gal}(M/K)$, then by Galois theory, $K$ corresponds to $G$ and $L$ corresponds to $G=\operatorname{Gal}(M/L)$ such that $|G/H|=p$, which might not be a group since $H$ might not be normal. But since $|G/H|=p$, there can't be any subgroup in between $G$ and $H$ and hence there can't be any subfields in between $L$ and $K$. Hence any element $\alpha \in L$ but not in $K$ is a primitive element.
But I am not sure if my proof is correct and also if there is any other way to prove it. Thanks.