The following claim is made:
Let $K$ be a field and $L$ the splitting field of a separable polynomial of degree $n \in \mathbb N$. Denote the zeros of $f$ in $L$ by $\alpha_1, \dots \alpha_n$. Then for $i =1,2, 3 , \dots, n$ $$[K(\alpha_1, \dots \alpha_i): K] \leq n (n-1) \dots (n-i+1)$$
Why is this true? I have seen cases where we have $n!$, but I suppose there is some $\frac{n!}{i!}$ going on here, what result is used here? Also does the user mean $L=K(\alpha_1, \dots \alpha_n)$ here?
I suppose one has to think about the degrees of the extensions: $[K(\alpha_1): K], [K(\alpha_1, \alpha_2): K]$, which gives the ascending chain $K \subset K(\alpha_1) \subset K(\alpha_1, \alpha_2) \dots $ but I do not know how to formalise this idea or make it useful.
EDIT: My attempt at a proof:
We indeed consider $K \subset K(\alpha_1) \subset K(\alpha_1, \alpha_2) \dots $ and their degrees. We will use the tower relation for fields. First of all we observe that $[K(\alpha_1):K]\leq n $. This is because when we adjoin a single root of a degree $n$ (minimal) polynomial to the field, the elements in the extension $K(\alpha_1)$ are linear combinations of $\{1, \alpha_1, \alpha_1 ^2 , \alpha_1 ^3 , \dots, \alpha_1 ^{n-1}\} $. Because $f$ is not given to be a minimal polynomial of $\alpha_1$, the actual minimal polynomial might be of lower degree than $n$ and we will not get all the powers listed before. This is why $[K(\alpha_1):K]\leq n $.. Now since $\alpha_1$ is a root of $f$, we have that $(X- \alpha_1) | f(X)$ and hence we know that $g(X):=\frac{f(X)}{X-\alpha_1}$ is of degree at most $n-1$. This new polynomial $g(X)$ is still zero at the remaining roots, we can repeat this process till we run out of roots. By the tower theorem we have: $$ [K(\alpha_1, \alpha_2):K] = [K(\alpha_1):K] \cdot [K(\alpha_1, \alpha_2):K(\alpha_1)] \leq n \cdot (n-1) $$ Continuing the argument above $i-2$ more times gives: $$ [K(\alpha_1, \alpha_2, \dots, \alpha_i):K] = \leq n \cdot (n-1)\dots (n-i+1) $$ Which is the desired inequality $\square$.