Given a number fields extension $F\subseteq K$ I'm trying to prove that the characteristic polynomial of an element $\alpha\in K$ - which is defined to be the characteristic polynomial of the $F$-endomorphism of $K$ given by the multiplication $x\mapsto \alpha x$ - is a power of its minimal polynomial.
I know that if $F\subseteq K \subseteq E$ is the Galois closure of $K/F$ then the minimal polynomial of $\alpha$ is $$ f^{min}_\alpha (x)=\prod_{\sigma\in Gal(E/F)}(x-\sigma(\alpha))$$ while if $\text{Hom}_F(K,\mathbb{C})=\{\sigma_1, \dots, \sigma_n\}$ its characteristic polynomial is given by
$$ f^{char}_\alpha (x)=\prod_{i=1}^n (x-\sigma_i(\alpha))$$
The result is the following $$ f^{char}_\alpha (x)=f^{min}_\alpha(x)^{[K/F(\alpha)]}$$
and the textbook proves it as follows. The multiplicity of the factor $(x-\sigma_i(\alpha))$ in the characteristic polynomial is given by the number of $\sigma_j$ s.t. $\sigma_j(\alpha)=\sigma_i(\alpha)$ which is the number of $F$-homomorphism $K\to E$ extending $\sigma_i\big|_{F(\alpha)}$.
How do I prove that number is given by $[K/F(\alpha)]$? If $K/F$ was Galois I would use the fundamental theorem, but in this general case?