How $N(2\zeta^{2n})=2^{p-1}$?

Question

Let, $\zeta$ be the $p$ th root of unity and $\mathbb{Z}[\zeta])$ be the number ring generated by $\zeta$, and $N$ is the norm function.

Why or how $N(2\zeta^{2n})=2^{p-1}$?

The source of the problem is -

Answer 1

The norm is a homomorphism from the multiplicative group $\mathbb Q[\zeta]^\ast$ to $\mathbb Q^\ast$,

so $N(2\zeta^{2n})=N(2)N(\zeta^{2n}),$ and $N(2)=2^{p-1}$ and $N(\zeta^{2n})=1$.

$[\mathbb Q[\zeta]:\mathbb Q]=p-1$, and $N(2)$ is the product of the $p-1$ conjugates of $2$, which are all $2$,

since $2$ is in the base field. $\zeta^n$ is a unit, so its norm is $\pm1$, so $N(\zeta^{2n})=1.$