This was part of a two part question which first asked to show that $[L : K]_s= [K_s: K],$ where $K_s$ is the separable closure of $K$ in $L$. I use this result in my answer. I'm looking for help showing \begin{equation*} [L:K]=p^k [L:K]_s \end{equation*} where $k$ is a non-negative integer. I've attached my attempt at a proof below. I would appreciate if you can point out where it goes wrong, or advise how it could be repaired. Thanks in advance.
We assume $K$ has order $p^n.$ Let $|L|=p^j$ and $|K_s|=p^i.$ Since $K$ is a subfield of $K_s$ which is a subfield of $L,$ we know that $n \mid i$ and $i \mid j$ i.e. $i=nx$ and $j=iy$ for $x,y \in \mathbb{N}.$ By the tower law for finite extensions, \begin{equation*} [K_s:K]=\frac{[L:K]}{[L:K_s]}. \end{equation*} Combining this with $[L:K]_s=[K_s:K]$ yields \begin{equation*} [L:K]=[L:K_s]\cdot [L:K]_s. \end{equation*} To continue, it seems necessary to compute $[L:K_s]$. I note that again by the tower law, \begin{equation*} [L:K_s]=\frac{[L:K]}{[K_s:K]}. \end{equation*} I get stuck now on how to compute this term. My approach was that if $\mathbb{F}_p$ was the base field, then \begin{equation*} [K_s:K]=\frac{[K_s:\mathbb{F}_p]}{[K:\mathbb{F}_p]}=\frac{i}{n}=\frac{nx}{n}=x. \end{equation*} Trying the same for $[L:K]$ gives \begin{equation*} [L:K]=\frac{[L:\mathbb{F}_p]}{[K:\mathbb{F}_p]}=\frac{j}{n}=\frac{nxy}{n}=xy. \end{equation*} Therefore, $[L:K_s]=\frac{xy}{x}=y$ however as I can't seem to prove that this $y$ is a power of $p$ I assume I've gone wrong somewhere.
Your argument leading up to showing the equivalence between the claim in the title and $ [L : K_s] $ being a power of $ p $ is correct. From there, you know that $ L/K_s $ is a purely inseparable extension, so it becomes a matter of knowing what are the possible degrees for such extensions.
You can show by a simple induction argument that a finite purely inseparable extension $ N(\alpha)/N $ with a primitive element $ \alpha $ must have degree a power of $ p $, where $ p = \operatorname{char} N $. It's obviously true if the extension is trivial, which gives the base case for the induction. In that case, since $ \alpha $ is inseparable over $ N $, its minimal polynomial $ f \in N[X] $ must be of the form $ g(X^p) $ for some $ g \in N[X] $. $ \alpha $ is a root of the polynomial $ X^p - \alpha^p $ in $ N(\alpha^p)[X] $, which is irreducible since otherwise we would have $ \alpha \in N(\alpha^p) $, so $ [N(\alpha) : N(\alpha^p)] = p $. We know by the induction hypothesis that $ [N(\alpha^p) : N] $ has degree a power of $ p $, so this shows by the tower law that $ [N(\alpha) : N] $ is a power of $ N $ as well.
For general finite purely inseparable extensions $ M/N $, you just pick any element $ \alpha \in M $ with degree $ > 1 $ over $ N $. $ [N(\alpha) : N] $ is a power of $ p $ by the above argument, and $ M/N(\alpha) $ is still purely inseparable of lower degree, so another induction argument along with the tower law finishes the proof.