Let $f(x)$ be irreducible in $F[x]$, $F$ of characteristic $p>0$. Show that $f(x)$ can be written as $g(x^{p^e})$ where $g(x)$ is irreducible and separable. Use this to show that every root of $f(x)$ has the same multiplicity $p^e$ (in a splitting field).
I have been able to prove that $f$ can be expressed in terms of $g(x^{p^e})$, where $g$ is irreducible and separable. Also what is the relation between the splitting field of $g(x)$ over $F$ and the splitting field of $f(x)$ over $F$?
Hint: Prove that for all $h\in F[x]$ you have $h(x^{p^e})=\operatorname{Frob}^eh(x)^{p^e}$.
Details: For every prime number $p$ and every positive integer $e$ the binomial coefficient $\tbinom{p^e}{k}$ is divisible by $p$ for all integers $k$ with $1<k<p^e$. It follows that for all $c\in\Bbb{F}_{p^e}$ and all $r\in\Bbb{F}_{p^e}[x]$ you have $$(cx^n+r(x))^{p^e}=c^{p^e}(x^n)^{p^e}+r(x)^{p^e}=\operatorname{Frob}^e(c)(x^{p^e})^n+r(x)^{p^e}.$$ By induction on the degree of $h$ it follows that $h(x)^{p^e}=\operatorname{Frob}^eh(x^{p^e})$ for all $h\in\Bbb{F}_{p^e}[x]$. In particular, this shows that every root of $h(x^{p^e})$ has multiplicity $p^e$, or a multiple thereof, and if $g\in F[x]$ is separable then every root of $g(x^{p^e})$ has the same multiplicity $p^e$.