I'm trying to solve the last part of an exercise in Dummit and Foote.
Let $K/F$ be any finite separable extension, and let $\alpha\in K$. Let $L$ be a Galois extension of $F$ containing $K$ and let $H\leq \text{Gal}(L/F)$ be the subgroup corresponding to $K$. Define the norm of $\alpha$ from $K$ to $F$ to be $ N_{K/F}(\alpha)=\prod_\sigma\sigma(\alpha), $ where the product is taken over all the embeddings of $K$ into an algebraic closure of $F$, (so over a set of coset representatives for $H$ in $\text{Gal}(L/F)$ by the Fundamental Theorem of Galois Theory.
The part I'm stuck on is
(d) Let $m_\alpha(x)=x^d+a_{d-1}x^{d-1}+\cdots+a_1x+a_0\in F[x]$ be the minimal polynomial for $\alpha\in K$ over $F$. Let $n=[K:F]$. Prove that $d$ divides $n$, that there are $d$ distinct Galois conjugates of $\alpha$ which are all repeated $n/d$ times in the product above and conlcude that $N_{K/F}(\alpha)=(-1)^na_0^{n/d}$.
I see that $d$ divides $n$, since $d=[F(\alpha):F]$, and this divides $[K:F]$ by multiplicativity of the degree. I also see that there are $d$ distinct Galois conjugates of $\alpha$ since the minimal polynomial is irreducible and has no repeated roots since $K/F$ is a separable extension. But why are they repeated $n/d$ times? After that, I think I can make the final conclusion.
I let H' be the subgroup corresponding to $F(\alpha)$, so H\leq H'\leq\text{Gal}(L/F). I calculate that [H':H]=n/d, but does this imply the result in some way?