TITLE: Multiplicative nature of the separability degree
QUESTION [8 upvotes]: In what follows, let $E / F$ be an algebraic extension, $h(x),f(x)\in F[x]$ polynomials, $h(x)$ irreducible. 
Definitions.

We say $h(x)$ is separable if it has not repeated factors. We say $f(x)$ is separable if all its irreducible factors are separable.
We say $\alpha\in E$ is separable over $F$, if its minimal polynomial over $F$ is.
The extension $E/F$ is separable if all the elements of $E$ are separable over $F$.
We define the field $$\newcommand\sclo{\operatorname{Scl}}\sclo_F E= \{\alpha\in E : \text{ $\alpha$ is separable over $F$ }\}$$ as the separable closure of $F$ in $E$.
The separability degree of the extension $E/F$ is defined as $$[E:F]_s = [\sclo_F E:F].$$

Now, suppose the extension $E/F$ is finite.
My question is: how can one proof that with these definitions, if $B$ is an intermediate field of the extension $E / F$, then
$$[E:F]_s = [E:B]_s[B:F]_s?$$
I know that the purpose of the separability degree is to count in how many ways we can extend certain injective homomorphism to an automorphism. If one starts defining separability degree this way, as done in the book by Lang, the result follows without much difficulty.
I wonder how to achieve it in this way.

REPLY [4 votes]: As I recall, when I taught Galois Theory, my practice was to define a concept (like separability of elements and extensions) and immediately give conditions that were equivalent to the definition. In that way, I could go back and forth among the various characterizations so as to make proofs of related theorems as quick and comprehensible as I could. But then, I was a disciple of Lang…