Let $(K_n)_{n\geq0}$ be a sequence of CM-fields, so that $K_0\subset K_1\subset\dots$ with $[K_{n+1}:K_n]=p$ for all $n\geq0$. For $n\geq0$ let $W_n$ be the group of the roots of unity in $K_n$. Now suppose that $\sigma$ is a generator of $\mbox{Gal}(K_{n+1}/K_n)$.
I want to show that $W^{\sigma-1}_{n+1}\subseteq W_{n+1}\cap \ker(N)$ where $N$ ist the norm map $N_{K_{n+1}/K_n}$ and $W^{\sigma-1}_{n+1}=\{\zeta^{\sigma-1}=\frac{\sigma(\zeta)}{\zeta}|~\zeta\in W_{n+1}\}$.
So I take $x\in W_{n+1}^{\sigma-1}$. Then there exists a root of unity $\zeta\in W_{n+1}$ so that $x=\frac{\sigma(\zeta)}{\zeta}$. Since $\sigma$ and "dividing by the identity" are Galois automorphisms, both $\sigma(\zeta)$ and $\zeta^{-1}$ are roots of unity so that $x\in W_{n+1}$.
Now compute the norm: $N(x)=N(\frac{\sigma(\zeta)}{\zeta})=N(\sigma(\zeta))N(\zeta)^{-1}=\prod\limits_{\tau\in Gal(K_{n+1}/K_n)} \tau(\sigma(\zeta))\prod\limits_{\tau\in Gal(K_{n+1}/K_n)} \tau(\zeta)^{-1}$. So how do I show that $N(x)=1$?
Since $\mbox{Gal}(K_{n+1}/K_n)$ is a group, the composition $\omega=\tau\circ\sigma$ of $\tau,\sigma\in \mbox{Gal}(K_{n+1}/K_n)$ again is a Galois automorphism and the following equation holds:
$N(x)=N(\frac{\zeta^\sigma}{\zeta})=\prod\limits_{\tau} \tau(\sigma(\zeta))\prod\limits_{\tau} \tau(\zeta)^{-1}=\prod\limits_{\omega} \omega(\zeta)(\prod\limits_{\tau} \tau(\zeta))^{-1}=N(\zeta) N(\zeta)^{-1}=1$.
Another way to show this is to recognize that the image of $N$ is a subset of $K_n$. So each $\sigma\in \mbox{Gal}(K_{n+1}/K_n)$ leaves $N(x)$ for all $x\in K_{n+1}$ constant, especially $N(\zeta)$ for $\zeta\in W_{n+1}\subset K_{n+1}$. It follows:
$N(x)=N(\zeta^{\sigma-1})=N(\zeta)^\sigma N(\zeta)^{-1}=\sigma(N(\zeta))N(\zeta)^{-1}=N(\zeta)N(\zeta)^{-1}=1.$