TITLE: Separable Polynomial Induction Proof
QUESTION [1 upvotes]: The following definition and lemmas are found in an undergraduate elliptic curves course.
Definition: Let $g(x) \in K[x]$ be a polynomial. We say that $g(x)$ is separable if $g'(x)$ is not identically zero. We say $g(x)$ is inseparable otherwise. 
Lemma 2: Let $K$ be a field of characteristic $p > 0$. Then,  $g(x) \in K[x]$ is inseparable if an only if there is a separable polynomial $t(x) \in K[x]$ such that $g(x)=t(x^{p^k})$ for some positive integer $k$.
I am told that this follows inductively from the following lemma:
Lemma 1: Let $K$ be a field of characteristic $p > 0$, and $g(x) \in K[x]$. Then, $g(x)$ is inseparable if and only if it is of the form $g(x) = h(x^p)$ with $h(x) \in K[x].$
I fully understand lemma 1. I do not see how lemma 2 follows inductively. Why do we get a separable polynomial $t(x)$ and why is $p$ raised to a power $k$? I am confident that the lemmas are correct given that I have found them stated elsewhere. I would appreciate any help! 
Thanks.

REPLY [1 votes]: If $g$ is inseparable, then $g(x)=h_1(x^p)$ and $h_1$ has degree strictly less than $g$. If $h_1$ is still inseparable, we may continue the procedure (with $h_1$ in place of $g$) until we end up with a polynomial $h_n$ that is not of the form $h_n(x)=h_{n+1}(x^p)$. But this means that $h_n$ is separable and $g=h_1(x^p)=h_2(x^{p^2})=\dots=h_n(x^{p^n})$. The converse is clear.
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