The Problem: Let $k \subset F \subset L$ such that $[L:k] < \infty.$ Let $S_1$ be the separable closure of $k$ in $F$, $S_2$ the separable closure of $F$ in $L$ and $S$ be the separable closure of $k$ in $L.$ Show that $[S:S_1]=[S_2:F]$ and $[F:S_1]=[S_2:S].$
My approach: Actually, I have tried only that $p=[S_1:k]$ and $q=[S_2:F]$ is the number of distinct $k$ and $F$-embeddings of $F$ and $L$ in $\bar k$ and $\bar F$ respectively. So, $q \geq p$ and then $r=[S:k]$.
Now, in both cases, I am unable to use any other fact to show the equality.
A help is warmly appreciated, thanks in advance.
First reduce to the question of showing that if $k \subset F$ is purely inseparable and $F \subset L$ is separable and $S$ is the separable closure of $k$ in $L$ then $[S:k] = [L:F]$, and the answer to this question is that they are both the number of $k$-embeddings $L \to \overline{k}$.
Or, more directly, show that the number of $k$-embeddings $L \to \overline{k}$ (under your original hypothesis) is equal to the number of $k$-embeddings $F \to \overline{K}$ multiplied by the number of $F$-embeddings $L \to \overline{k}$.