I do not understand the underlined equality in the following context:
As I can see, one tried to apply the map $B$ on $F[B(E)^{1/n}]$ and $F$. I can see that the equality makes sense if we know that $B$ is a bijection. However, I think this is what we are trying to show in this theorem, so this argument does not work.
Could you please help explaining to me this equality?
Let $E/F$ be a finite Galois extension with Galois group $G$. On page $72$ of Fields and Galois Theory, Milne shows that $F^{\times}\cap E^{\times n}/F^{\times n}$ is isomorphic to ${\rm Hom}(G,\mu_n)$ and if $G$ is abelian of exponent $n$, $$|{\rm Hom}(G,\mu_n)|=(G:1)=[E:F].$$ Therefore if $G$ is abelian of exponent $n$, $$[E:F]=(F^{\times}\cap E^{\times n}:F^{\times n})=(B(E):F^{\times n}),$$ where $B(E)=F^{\times}\cap E^{\times n}$.
In your case, $F[B(E)^{1/n}]$ is a finite abelian extension of exponent $n$ (actually, Milne proves this in the next paragraph) and the equality follows.