$\lvert\operatorname{Aut}(F/K)\rvert$ divides $[F:K]$ Let $F/K$ be a finite extension.
I want to prove $\lvert\operatorname{Aut}(F/K)\rvert$ divides  $[F:K]$.
I know  $\lvert\operatorname{Aut}(F/K)\rvert\leq [F:K]$ but I cannot show the former.

Take $\alpha$ from $K$, then $\lvert\operatorname{Aut}(F/K)\rvert＝\lvert\operatorname{Root}P_{\alpha}(K)$.
$[L:K]＝\deg P_{\alpha}$.
But I cannot go further. Thank you in advance for helping me.
Denote $G=\mathrm{Aut}_K(F)$. Then $F/F^G$ is a Galois extension with Galois group $G$. Thus, we have
$$|G|=[F\colon F^G]=\frac{[F\colon K]}{[F^G\colon K]}\Big\vert[F\colon K].$$
Every automorphism of $F$ over $K$ is determined by its action on a basis, which has $[F\colon K]$ elements. Thus, if the basis is mapped into itself by the automorphism group, then Aut$(F/K)$ can be represented as a subgroup of the permutation group on $[F\colon K]$ elements. Thus, the number of elements of Aut$(F/K)$ divides the size of the permutation group, which is $[F\colon K]$ factorial.

Your question asks about a much smaller bound, without the factorials. And for this we need a more refined approach. The way that the inequality $|\textrm{Aut}(F/K)|\leq [F\colon K]$ is proven here is to decompose the finite extension into a tower of simple extensions. Both sides of the inequality are multiplicative, and for simple extensions it is straightforward to get the bound.

Now if you follow along in this proof, each time you obtain an inequality between two integers, it is actually showing that one divides the other. Hence your result follows.

An earlier version of this answer was incorrect. The example I was considering was the following: a cubic extension with automorphism group $S_3$, which I believed could be obtained from an explicit construction in the comments below. That construction is incorrect, and no such extension exists.