$\lvert K(\alpha_1,\dots,\alpha_n) : K \rvert$ is a divisor of $n!$

Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n>0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:

$$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$

I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.

The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.

Any idea for solving this problem would be appreciated.

The base case $n=1$ is clear. If $n >1$: suppose that $f(x)$ is irreducible. Then $[K(\alpha_1): K] = n$, so by the tower theorem it remains to show that the degree of $[K(\alpha_1,..., \alpha_{n-1}, \alpha_n): K(\alpha_1)]$ divides $(n-1)!$. Of course we already have this by induction, since here we are adjoining the roots of a degree $n-1$ (or less) polynomial to $K(\alpha_1)$ and our base case works over any field.

Suppose that $f$ splits into $k$ irreducible factors over $K$, $f(x) = p_1(x)\cdots p_k(x)$, of respective degrees $d_1,..., d_k < n$, $d_1 +\cdots +d_k = n$. By induction we know that the splitting field of $p_i(x)$ has degree over $k$ dividing $d_i!$. Therefore the degree over $K$ of the field obtained by adjoining all roots of $f$ to $K$ divides $d_1!\cdots d_k!$ (simply adjoin the roots of $p_1(x)$, then $p_2(x)$, etc. and apply the inductive hypothesis and the tower theorem), which divides $n!$ by the fact that the multinomial coefficient is an integer.

Let $p(x)$ be an irreducible factor of $f(x)$. In the field $$L = \frac{K[x]}{(p(x))},$$ the coset $\overline{x}$ is a root of $p(x)$ and $[L : K] = \textrm{deg}(p) \leq n$. Proceed by induction until you have produced a field $M$ that contains all roots of $f(x)$. Since you are adjoining one root at a time the upper bound on the degree of the extension drops by $1$ each time, thus $[M : K] \leq n!$, so $[M : K]$ divides $n!$.

Suppose $f$ is reducible over $K$ but has no root in $K$. Then $f(x) = g(x)h(x)$ for some $g(x),h(x) \in K[x]$ with $a = \textrm{deg}(g), n - a= \textrm{deg}(h) \geq 2$. Then $$[M : K] \leq a!(n - a)! < (n - 1)!$$ as long as $n \geq 3$. If $n = 1$ the statement is trivial and if $n = 2$ the statement is easily seen to be true because in that case $f(x)$ is reducible over $K$ if and only if $f(x)$ has a root in $K$.