Theorem Let $F$ be a field. Let $K$ be a finite extension of $F$. Let $[K : F]_i$ be the inseparable degree of $K/F$. Let $\bar{K}$ be an algebraic closure of $K$. Let $S$ be the set of $F$-embeddings of $K$ into $\bar{K}$. Let $\alpha \in K$. Then $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$
This is proved in Theorem 60 in page 39 of the lecture note written by Pete L. Clark. I don't understand the proof. Would any one please enlighten me?
EDIT Why down votes?? Asking for help to understand a proof should be frowned upon?
EDIT I understand the prerequisites for the proof, i.e. the content of section 6 and Corollary 58.
EDIT Related question 1, Related question 2
EDIT Let $f(X)$ be the characteristic polynomial of $\alpha$. Let $g(X)$ be the minimal polynomial of $\alpha$. $By$ $ $ $Corollary$ $ $ 58, $f(X)$ = $g(X)^{[K:F(\alpha)]}$. The set {$\sigma(\alpha); \sigma ∈ S$} is the set of the roots of g(X). However, it is not clear to me that the equation $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$ follows immediately. Would anyone please explain why the equation follows other than just saying it is straightforward?
EDIT It's amazing that some people regard questioning a proof as an attack to the author's credibilty. Everybody makes a mistake. Even Grothendieck made a non-trivial mistake(see Misconceptions About $K_{X}$ by Kleiman). I think this attitude is harmful to healthy development of mathematics. I'm not claiming that the proof is wrong, though.
EDIT To anyone who thinks the proof is straightforward, please explain to me in detail. I am not as smart as you.
EDIT It's surprising that no one explained the proof so far. I believe every step of any proof can be reduced to (really) trivial statements. If you think it's straightforward, please reduce it to more trivial statements that anybody who has basic knowledge of abstract algebra can understand.