For a field $K$ of characteristic $p > 0$, and a finite field extension $L/K$, let $K_s$ be the separable closure of $K$ in $L$. I am to show that
$[L : K]_s = [K_s : K]$.
This would mean that the number of embeddings of $L$ in an algebraic closure of $K$ (LHS) equals the number of basis vectors in $K_s$ as a vector space over $K$ (RHS). As $L/K$ is finite, every subextension of $K$ is finite, so $K_s/K$ is finite. It is also separable by construction, and is hence, for some $t ∈ ℕ$ and separable $\alpha_1, \alpha_2, \ldots, \alpha_t ∈ K_s$, given by
$$K_s = K(\alpha_1, \alpha_2, \ldots, \alpha_t) = K(\alpha_1, \alpha_2, \ldots, \alpha_{t-1})(\alpha_t) = \ldots = (\ldots(K(\alpha_1))(\alpha_2) \ldots)(\alpha_t).
$$
In fact, finite separable extensions have a primitive element: $K_s = K(x)$ for some $K$-separable $x$, so yet another way to express the RHS is
$$[K_s : K] = \deg f^x_K.
$$
Moreover, by the separability of the $\alpha_i$, their respective minimal polynomials $f^{\alpha_i}_{K(\alpha_1, \alpha_2, \ldots, \alpha_{i-1})} ∈ K(\alpha_1, \alpha_2, \ldots, \alpha_{i-1})[X]$ have exactly as many roots as their degree, so we could write
$$ [K_s : K] = \prod\limits_{i = 1}^t \deg f^{\alpha_i}_{K(\alpha_1, \alpha_2, \ldots, \alpha_{i-1})}.
$$
Lastly, by the separability and finiteness of $K_s/K$, we have $[K_s : K] = [K_s : K]_s$.
You can see that I have some knowledge about the structure and dependencies in this world of separable extensions, but it is very much unconnected and unorganised. I feel that all ingredients should be on the table now, but that I lack a key insight that will connect all of this together.
PS: One might wonder where the characteristic of $K$ comes in, but I think that only pertains to the second part of the question: show that $(\exists k ∈ ℕ)([L : K]_s · p^k = [L : K])$.
Best Answer
Given that you seem to have some basic understanding of the material, you might be able to prove that $[L:K_s]_s=1$ and then use the multiplicativity of the degree of seperability to deduce that $$[L:K]_s=[L:K_s]_s\cdot [K_s:K]_s=[K_s:K]$$
The assumption that $K$ has positive characteristic will certainly be used in the second part of the exercise. Another nice fact: Any field of characteristic 0 is "perfect", i.e. any finite extension is seperable, so we would have nothing to show here.