I am puzzled with the following (apparently easy ?) question. Let $k$ be a field of characteristic $p>0$, such that $[k:k^p]<\infty$ (in other words, $k$ is $F$-finite, with $F: k\rightarrow k$ the Frobenius morphism).
Denote $k^s$ the separable closure of $k$: I want to show that $(k^s)^p$ generates $k^s$ over $k$. To do so, considering $L/k$ a finite separable extension of degree $n$, I'd like to show that $L^p$ generates $L$ over $k$, i.e., that if $e_1$, ..., $e_n$ is a $k$-basis of $L$, then $e_1^p$, ..., $e_n^p$ is still a $k$-basis of $L$. Writing $L=k(x)$ for some $x\in L$ and taking $e_1=1$, $e_2=x$, ..., $e_n=x^{n-1}$, this is equivalent to showing that $1$, $x^p$, ..., $x^{p(n-1)}$ remains free over $k$.
EDIT: By analyzing the dimensions, assuming that the family $1$, $x^p$, ..., $x^{p(n-1)}$ is not free over $k$ would imply that $n$ is divisible by some $p^m$ (with $p^m$ the degree of the finite radical extension $L/k(x^p)$). If $n$ is prime to $p$ then we're fine.
EDIT2: In fact, the extension $L/k(x^p)$ is necessarily trivial. As $L/k$ is separable, then $L/k(x^p)$ is also separable (the minimal polynomial of any element of $L$ over $k(x^p)$ divides its minimal polynomial over $k$, hence has distinct roots): then $L/k(x^p)$ is trivial, for it's also radical. Thanks MooS for pointing it out.
So we didn't need the condition $[k:k^p]<\infty$.