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

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