Every finite separable extension is contained in a Galois extension

I am having trouble understanding the following proof:

Claim: Let $$K/F$$ be a finite separable field extension. Then $$K$$ is contained in a Galois extension $$K \supset L \supset F$$.

Proof: Since the extension $$K/F$$ is finite, we can write $$K=F(\alpha_1, \dots, \alpha_k)$$ for some $$\alpha_i$$. If $$f_i$$ is the minimal polynomial of $$\alpha_i$$ then one can choose $$L$$ to be the splitting field of the polynomial $$f=f_1\dots f_n$$, then $$L\supset K\supset F$$ and $$L\supset F$$ is Galois.

Why is the polynomial $$f$$ separable?

If you just multiply all the minimal polynomials then it might not be separable, because maybe two elements $$\alpha_i,\alpha_j$$ have the same minimal polynomial. But you can take the distinct minimal polynomials of the elements $$\alpha_1,...,\alpha_k$$, say $$f_1, f_2,...,f_s$$ and then take the splitting field of $$f=f_1f_2...f_s$$. This splitting field is the same as the splitting field of the polynomial you have defined, the only difference is that this polynomial is indeed separable. (which proves the extension is Galois). It is separable because of the following two facts:
$$1$$. All the polynomials $$f_1,...,f_s$$ are separable because $$K/F$$ is a separable extension.
$$2.$$ If $$i\ne j$$ then $$f_i$$ and $$f_j$$ have no common roots in the algebraic closure. This is because two distinct monic irreducible polynomials are coprime. And it is a standard result that if two polynomials are coprime over $$F$$ then they are coprime over any extension of $$F$$, including the algebraic closure.
Combine these two facts, it clearly follows that $$f$$ can't have a multiple root.