QUESTION: Suppose $\zeta$ is an primitive n-th root of unity. And suppose $n=p^r$ where $r\geq 1$. What is $(1-\zeta)\zeta^{p^r-p^{r-1}-1}$ written as the sum of the basis elements $1,\zeta,\zeta^2,...,\zeta^{p^r-p^{r-1}-1}$. I've been trying to find an identity so that I can reduce this, but I seem to be stuck. Any suggestions.
As a side note I'm basically trying to calculate $N_{K\backslash\mathbb{Q}}(1-\zeta)$, where $N$ is the norm and $K=\mathbb{Q}(\zeta)$.
We have $$ N_{K/\mathbb{Q}}(1-\zeta)=\prod_{(j,p^r)=1}(1-\zeta^j)=\Phi_{p^r}(1)=p, $$ where $\Phi_n(t)$ denotes the cyclotomic polynomial. For this you need that the norm is the product over the Galois conjugates, and that $Gal(K/\mathbb{Q})\simeq (\mathbb{Z}/p^r)^{\times}$, the Galois conjugates just being the $\zeta^j$'s. And perhaps you would like to see that $$ \Phi_{p^r}(t)=t^{p^{r-1}(p-1)}+\cdots +t^{p^{r-1}}+1, $$ so that $\Phi_{p^r}(1)=p$.
To your first question, just multiply out $(1-\zeta)\zeta^m=\zeta^m-\zeta^{m+1}$.
The discriminant is given by $$ D(1,\zeta,\zeta^2,\ldots ,\zeta^{\phi(p^r)-1})=\pm p^{p^{r-1}(pr-r-1)}. $$