Given $F$, a field with characteristic $p > 0$, and a finite purely inseparable field extension $E$. Then prove that the minimal polynomial $f(x)$ of any $\alpha \in E\backslash F$ will be of the form $X^{p^n} - \alpha^{p^n}$, with $n \geq 0$ being an integer.
Now, I've tried to prove this, using Isaacs' "Algebra: A Graduate Course" as a guide. However, after a series of corollaries and theorems referring to at least one previous theorem or corollary every time, I think I have it. I'll try to put the proof here (more or less):
Isaacs proves that $f(x)$ must be of the form $X^{p^n} - \alpha^{p^n}$, because $f(x)$ can be written as $g(x^{p^n})$ with $g \in F[X]$ irreducible and separable over $F$. This is proven because we can write $f(x) = h(x^p)$ for some irreducible $h \in F[x]$, and then using induction on the degree.
That $f(x)$ can be written as $h(x^p)$ is in turn proven by using that $f' = 0$, then deriving $f$ and showing that coefficients $a_i$ must equal $0$ if $p\nmid i$.
Again, in turn is proven that $f' = 0$ because, since $f$ does not have distinct roots, there must be an $a$ in a certain field $K$ such that $(x-a)^2|f(x)$ in $K[x]$. Writing $f(x) = (x-a)^2h(x)$ gives that $f'(x) = 2(x-a)h(x) + (x-a)^2h'(x)$, which shows that $f'(a)=0$. Then, since $f$ is irreducible, it divides every polynomial in $F[x]$ that has $a$ as a root. Thus $f|f'$ but since $\deg f > \deg f'$, $f' = 0$.
I'm wondering if I made any mistakes/inconsistencies/... ?
And if there is really no easier or more straightforward way to prove this?