If $E|_k$ is a finite extension such that $E^pk=E$, then prove $E^{p^m}k=E$ where $p=$char$(k)$

Question

Here $E^p:=\{x^p:\ x\in E\}$. As $k$ is field of charateristic $p$, $E^p$ is a subfield of $E$.

Now given that $E^pk=E$. I'm trying to show that $E^{p^m}k=E$.

If I can prove that $E^pE^p=E^{2p}$. Then It will follow that $E^{2p}k=E^pE^pk=E^pE=E$. By induction $E^{np}k=E$, hence $E^{p^m}k=E$.

But I'm unable to prove $E^pE^p=E^{2p}$.

Then how to argue $E^{p^m}k=E$. This might look easy but I'm not getting it. Can anyone help me in this regard?

Thanks for help in advance.

Answer 1

Applying $(\_)^p$ $n$ times to $E^pk=E$ you get $E^{p^{n+1}}k^{p^{n}}=E^{p^n}$, then if you know that $E^{p^n}k=E$ you get $E^{p^{n+1}}k=E^{p^{n+1}}k^{p^n}k=E^{p^n}k=E$ (because $k=k^{p^n}k$). From this the claim follows by induction.