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 $[F(a,b):K]\leq [F(a)(b):F(a)][F(a):F]$
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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.
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1$\begingroup$ This answer contains the same proof as the one you just posted. $\endgroup$ May 9, 2016 at 14:05