Relation of compositum of fields Let $E/k$ be a finite field extension, $\operatorname{char}(k)=p>0$. Suppose that $E^p k = E$. Is it then true that $E^{p^n}k = E$ for any positive integer $n$? If yes, why?
Thanks.

Yes, it is true. I will show that $E=E^{p^2}k$ and leave to you the proof of the general case $E=E^{p^n}k$.
Since $E=E^{p}k$, any element $e\in E$ can be written as $e=\sum q_ie_i^p\;$ (for some $e_i\in k, q_i\in k$)   .
[This is due to the fact that the ring formed by the sums on the right is already a field, because that ring is a $k$-subalgebra of  the algebraic extension $E/k$]
In the same way, each  $e_i$ can be written as $e_i=\sum q_{ij}e_{ij}^p$.
Substituting yields $$e=\sum q_i(\sum q_{ij}e_{ij}^p)^p=\sum q_iq_{ij}^pe_{ij}^{p^2}$$
which shows that indeed $E=E^{p^2}k$
