Degree of splitting field divides n!

Question

If $f\in K[x]$ has degree $n$ over $f$, and $F$ is the splitting field of $f$ over $K$ then, show that $[F:K]\mid n!$

I can show that $[F:K] \leq n!$ using the fact that $[F(a,b):K]\leq [F(a)(b):F(a)][F(a):F]$

Answer 1

I am writing here the solution for the case $char K=0$. In this case, $F$ is a normal and hence a Galois extension of $K$. So, $|Gal(F/K)|=[F:K]$ and the Galois group can be thought of as a subgroup of $S_n$ which has order $n!$ and hence the result follows.