Prove the following product formulas for separability and inseparability degree: If $F\subseteq L\subseteq K$ are fields, then show that $[K : F]_s = [K: L]_s[L: F]_s$ and $[K: F]_i = [K: L]_i[L: F]_i$
This question has already been posted twice here show that $[K:F]_s=[K:L]_s[L:F]_s$ and $[K:F]_i=[K:L]_i[L:F]_i$. , and here $[K : F]_s = [K : L]_s [L : F]_s $ and $[K : F]_i = [K : L]_i [L : F]_i $, but in one they do not answer and in another I do not understand the answer because they use another definition for this grade. I am following another book and in that book they define it in the following way:
Let $K$ be a finite extension of $F$. If $S$ and $I$ are the separable and purely inseparable closures of $F$ in $K$, respectively, we define the separable degree $[K : F]_s$ of $K/ F$ to be $[S : F]$ and the insepaTable degree $[K : F]_i$ to be $[K : S]$. With these definitions, we see that $[K: F]_s[K : F]_i = [K : F]$.
I then have that $[K: L]_s[L: F]_s=[S_1:L][S_2:F]$ where $S_1$ and $S_2$ are the separable closure of $L$ in $K$ and $F$ in $L$ respectively, how can I relate this to $[K:F]_s=[S:F]$?