# Splitting field $L$ of polynomial $f \in K[x]$ with degree $n$ satisfies $[L:K] | n!$

Suppose $$f \in K[x]$$ is a polynomial with degree $$n$$, $$f = (x-\alpha_1)...(x-\alpha_n)$$ over the algebraic colsure. Let $$L=K(\alpha_1,...,\alpha_n)$$ be the splitting field of $$f$$. Prove that $$[L:K]$$ divides $$n!$$.
I was able to prove it for the case where $$f$$ is seperable. In this case, $$L/K$$ is a Galois extension and therefore $$[L:K] = |Gal(L/K)|$$ and since $$Gal(L/K)$$ is isomoprhic to a subgroup of $$S_n$$, the result follows. How do I prove it for the general case?

Proceed by induction on $$n = [L:K]$$. If $$[L:K] = 1$$ then the claim is trivial.
Assume $$[L:K] > 1$$. We consider separately the cases where $$f$$ is irreducible and where $$f$$ is not irreducible.
Suppose first that $$f$$ is irreducible. Let $$\alpha$$ be a root of $$f$$ in a splitting field. Then $$f$$ is the minimum polynomial of $$\alpha$$ over $$K$$, so $$[K(\alpha):K] = n$$. Then $$[L : K(\alpha)] < [L :K]$$ and $$L$$ is the splitting field of $$g(x) = f(x)/(x - \alpha)$$ over $$K(\alpha)$$. The degree of $$g$$ is $$n - 1$$, so by induction $$[L : K(\alpha)] \mid (n-1)!$$, and the result follows by the tower law.
Suppose now that $$f$$ is not irreducible. Then $$f = pg$$ for $$p, g \in K[x]$$ with $$p$$ irreducible. If $$L$$ is a splitting field for $$p$$ over $$K$$, then we are done by the previous paragraph. If not, we may take $$K \subset M \subset L$$ such that $$M$$ is a splitting field for $$p$$ over $$K$$ (just adjoin the roots of $$p$$ in L).
Then $$M$$ is a splitting field for $$p$$ over $$K$$, and $$L$$ is a splitting field for $$g$$ over $$M$$. Since our tower of fields has strict inclusions, $$[L:K] = [L:M][M:K]$$ is a proper factorisation, so by induction $$[L:M] \mid (\deg g)!$$ and $$[M :K] \mid (\deg p)!$$. If we define $$k = \deg p$$, then $$n - k = \deg g$$, so we have $$[L:K] = [L:M][M:K] \mid k!(n-k)!$$
And the result follows because $$k!(n-k)! \mid n!$$ (which is true because binomial coefficients are integers).