# Tower relation for field degrees and separable polynomial in splitting field

I have the following example exercise:

Let $$K$$ be a field and $$L$$ the splitting field of a separable polynomial $$f\in K[X]$$ of degree $$n$$. Denote the zeros of $$f$$ in $$L$$ by $$\alpha_1,\alpha_2,...,\alpha_n$$. Prove that for $$i=1,2,...,n$$: $$[K(\alpha_1,...,\alpha_i):K]\leq n(n-1)...(n-i+1).$$

I use induction and the tower relation for field degrees which says that for $$K\subset L\subset M$$ we have $$[M:K]=[M:L][L:K]$$.

My problem with the proof is that $$f$$ is not necessarily a minimum degree polynomial of the zeros (I think), which I struggle with right at the $$i=1$$ step of the induction, in which I try to work around this:

We look at the case $$i=1$$. We take $$g(x)=f(x)$$ if $$f(x)$$ is the polynomial of minimum degree such that $$f(\alpha_1)=0$$. Then $$\deg(g)=n$$. If $$f(x)$$ is not the minimum degree polynomial, then there exists $$q(x)\in K[x]$$ such that $$g(x)=\frac{f(x)}{q(x)}$$ is the minimum degree polynomial with $$g(\alpha_1)=0$$, where $$g(x)\in K[x]$$. Then $$\deg(g). Therefore taking both cases into account we have $$\deg(g)\leq n$$ and also $$\deg(\alpha_1)=\deg(g)\leq n$$. We then have that $$[K(\alpha_1) : K]=\deg(\alpha_1)\leq n$$.

I don't know if this is correct, because in a subsequent exercise I want to prove that $$L=K(\alpha_1,...\alpha_n)$$ divides $$n!$$. But because we might use the $$g$$ function with degree less than $$n$$ in the base case (and analogously with lesser degree in all subsequent steps), I think we might end up with for example $$[K(\alpha_1,...,\alpha_n):K]=n(n-1)(n-3)(n-3)(n-4)(n-5)...1$$ which does not divide $$n!$$.