A problem on separable extension on field of characteristic $p>0$ 
Problem: Let $k$ be a field of characteristic $p>0$. Let $\alpha $ be algebraic over $k$ then show that $\alpha$ is separable if and only if $k(\alpha)=k(\alpha^{p^n})$ for all positive integer $n$.

I have a solution below but I am not convinced if it works. Other approaches are always welcome.
Thank you.
 A: Suppose we have  $k(\alpha)=k(\alpha^{p^n})$ for all positive integer $n$.
We are to show that $\alpha$ is separable (i.e. the minimal polynomial of $\alpha$ over $k$ has no repeated roots in any extensions over $k$).
If possible let us assume that $\alpha$ is not separable. As it is a field of characteristic $p(>0)$ the minimal polynomial of $\alpha$ over $k$ will be $g(x^p)$ for some $g(x)\in k[x]$.
$\therefore g(\alpha^p)=0\implies g(x)$ annihilates $\alpha^p\implies$ $\textit{min-poly }_k(\alpha^p) $ divides $g(x)$
Now from  $k(\alpha)=k(\alpha^{p})$ we have $$[k(\alpha):k]=[k(\alpha^p):k]=\deg(\textit{min-poly }_k(\alpha^p) )\le\deg(g(x))< p\deg(g(x))=\deg(g(x^p))=[k(\alpha):k]$$
But that is a contradiction. So we must have $\alpha$ is separable.

Conversely, suppose $\alpha$ is separable we are to show that $k(\alpha)=k(\alpha^{p^n})$ for all positive integer $n$.
It is clear that for all positive integer $n$ , $\alpha^{p^n}\in k(\alpha)\implies k(\alpha^{p^n})\subseteq k(\alpha)$.
Let us fix $n$ a positive integer.
$f(x)$ be the minimal polynomial of $\alpha$ over $k$ and $g(x)$ be the minimal polynomial of $\alpha^{p^n}$ . Also $[k(\alpha^{p^n}):k]=\deg(g)\le \deg(f)=[k(\alpha):k]$
Let $L$ be the splitting field of $f$ over $k(\alpha)$.
Let us define $I:=\{ \sigma|\sigma:k(\alpha)\to L \text{ is a } k-\text{endomorphism} \}$. Observe that number of elements in $I$ is equal to $[k(\alpha):k]$, as $\alpha$ is seperable.
Moreover $\{\sigma(\alpha)|\sigma\in I\}$ is the set of $[k(\alpha):k]$ distinct roots of $f$ in $L$.
Then in $L$ polynomial $f(x)$ can be written as $$f(x)=\prod_{\sigma\in I}\left(x-\sigma(\alpha)\right)$$
Next we look at $g(x)$. $g(x)$ is an irreducible in $k[x]$. We have $\alpha^{p^n}\in L$ a root of $g(x)$,
then $g(x)$ splits in $L$.
For every $\sigma\in I$ , $\sigma(\alpha^{p^n})$ is a root of $g(x)$ and these roots are distinct  $(\because \sigma_1(\alpha^{p^n})=\sigma_2(\alpha^{p^n})\implies(\sigma_1(\alpha))^{p^n}=(\sigma_2(\alpha))^{p^n}\implies \sigma_1(\alpha)=\sigma_2(\alpha)\implies \sigma_1=\sigma_2)$.
So the $\deg(g)$ is greater or equal to the number of elements in $I$.
$\therefore \deg(g)=[k(\alpha):k]=[k(\alpha^{p^n}):k]\implies k(\alpha)=k(\alpha^{p^n})$
