Proposition about of Purely inseparable extension

Question

In the Book's Algebra, Lang, page 251.

Proposition: Let $E^p$ denote the field of all elementos $x^p$, $x\in E$- Let $E$ be a finite extension of $k$. If $E^pk=E$, then $E$ is separable over $k$.

Proof:Let $E_0$ be the maximal separable subfield of $E$. Let $E=k(\alpha_1,\ldots,\alpha_n)$. Since $E$ is purely inseparable over $E_0$ there exists $m$ such that $\alpha_i^{p^m}\in E_0$ for each $i=1,\ldots,n$. Hence $E^{p^m}\subset E_0$ . But $E^{p^m}k=E$ whence $E=E_0$ is separable over $k$.

$\textbf{My question's are:}$

1) Why can the same $m$ be used for all the $\alpha_i's$, rather than different $m_i's$?

2) Please explain why $E^pk=E$ implies $E^{p^m}k=E$ Thank you all.

Answer 1

For (1), there's some $m_i$ that works for each $\alpha_i$. It's enough to let $m$ be the largest of these.

For $(2)$, assume that $E^{p^{m-1}}k = E$. Apply the Frobenius endomorphism to each side of this equality to obtain $E^{p^m} k^p = E^p$, hence $E^{p^m} k = E^{p^m} k^p k = E^p k = E$.