Let $K=\mathbb{Q}[\xi_m]$ be a number field, where $\xi_m= e^{\frac{2\pi i}{m}}$ and $m=p^l$. I want to calculate the norm of $(1-\xi_m)$. I know how to do this if $l=1$, but if $l> 1$ I get a bit confused. This is my work so far:
First of all I know that $$N(1-\xi_m)=\prod_{1\leq k\leq m; (k,m)=1}(1-\xi_m^k)$$ I also know that $$\Phi_m(x)=\frac{x^m-1}{\prod_{d|m, d<m}\Phi_d(x)}=\frac{x^m-1}{\prod_{d|p^{e-1}}\Phi_d(x)}=\frac{x^{p^e}-1}{x^{p^{e-1}}-1}$$ However I don't know what to do with this. If $l=1$, then this is what I would do: $$N(1-\xi_p)=\prod_{k=1}^{p-1}(1-\xi_p^k)=\Phi_p(1)=p$$ I was hoping to do something similar in the case where $l\neq 1$, however even if I managed to prove that $N(1-\xi_m)=\Phi_m(1)$, I still wouldn't have my solution since I cannot tell the value of $\Phi_p(1)$ from the expression above (it would be $0/0$). Can someone help me?
As you say, $N(1-\xi_m)=\prod_{k\in(\mathbb Z/m)^\times}(1-\xi_m^k)=\Phi_m(1)$. Now there are several ways to evaluate this.
First of all, $\Phi_m(x)=\frac{x^{p^e}-1}{x^{p^{e-1}}-1}$, so using L'Hospital, $$\lim_{x\to 1}\Phi_m(x)=\lim_{x\to 1}\frac{(x^{p^e}-1)'}{(x^{p^{e-1}}-1)'}=p.$$ Alternatively, using geometric series, we have
$$\Phi_m(x)=\frac{(x^{p^{e-1}})^p-1}{x^{p^{e-1}}-1}=1+x^{p^{e-1}}+\dots+x^{p^{e-1}(p-1)},$$ from which it is immediate that $\Phi_m(x)=p$.
By the way, since $N(1-\xi_m)=p$ is prime, this shows $1-\xi_m\in\mathbb Z[\xi_m]$ is a prime element, lying above $p\in\mathbb Z$.