## Irreducible polynomial in a field of characteristic p

$\mathbf{Problem:}$ Let $f(x)$ be irreducible in $F[x]$. $F$ of charactersitic $p$. Show that $f(x)$ can be written as $\displaystyle g(x^{p^e})$ 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).
$\mathbf{Proof:}$ Suppose $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$. If some $m$ not divisible by $p$ such that $a_m\neq 0$, then $f'(x)\neq 0$ and $f$ is separable.
Otherwise $f(x)=b_nx^{np}+b_{n-1}x^{(n-1)p}+\cdots+b_1x^p+b_0=g(x^p)$ where $g(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0$. Since $\deg g<\deg f$, by induction, there is $h$ such that $\displaystyle f(x)=h(x^{p^e})$ and $h$ is separable.
Since $h$ is separable, it has simple roots in splitting field $E$. Suppose $h(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_s)$. Since $[E:F]<\infty$, $E$ is still a field of characteristc of $p$. Then $x^{p^e}-\alpha_i\in E[x]$ has a root of multiplicity $p^e$ in a splitting field over $E$. For different $i$, these roots are apparently different and they are all roots of $f$. So every root of $f$ has the same mutiplicity $p^e$.
$\mathbf{Remark:}$ Jacoboson, Algebra I. p234