Suppose, $F$ is a field of char $p>0$. Let $\alpha$ be algebraic over $F$ with minimal polynomial $f$. If $f$ is not separable, it can be shown that all roots (say we have $r$ of them) of $f$ have the same multiplicity, say $m$. Then, there exists a separable polynomial $h$ and an integer $n$ such that $h(x^{p^n})=f(x)$. Now, I want to show, for every root $\beta$ of $f$, $\beta ^{p^n}$ is a root of $h$ and every root of $h$ arises this way. The first part is quite clear, but I am unable to see the second part. I should be able to show this by a counting argument and comparing degrees if I could prove that $m=p^n$, but I am unable to see this either. I tried equating degrees of extensions involved and using the fact that the separable degree of $F(\alpha ^{p^n})$ over $F$ is equal to the degree of this extension. However, I am still unable to show this. Any hints?
PS: Feel free to edit the title if it's not very descriptive.