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Assume $T\in\mathbb{C}$, $m>1$ and $\mu_i$ be the $m$-th roots of unity. I want to prove that \begin{align} \prod_{i=1}^m \left(1-\mu_i T\right)=1-T^m. \end{align} By the brute force ansatz things get a little bit complicated. If I just expand the product I get \begin{align} \prod_{i=1}^m \left(1-\mu_i T\right)&=1-\left(\sum_{i=1}^m \mu_i\right)T+\left(\sum_{i_1<i_2}^m \mu_{i_1}\mu_{i_2}\right)T^2-...+(-1)^m \left(\mu_1\cdots \mu_m\right)T^m \\ \end{align} or \begin{align} &=1-T^m+\sum_{k=1}^{m-1}(-1)^k\!\!\!\!\!\!\!\!\!\sum_{1\leq i_1<...<i_k\leq m} \!\!\!\!\!\!\!\! \mu_{i_1}\cdots\mu_{i_k} T^k \end{align} Okay, now to proof that the last sum has to vanishes, I need some generalized orthogonality relation. For the primitive root $\mu_0$ we set $\mu_{i_j}=\mu_0^{i_j}=\exp\left(2 \pi i \frac{i_j}{m}\right)$. \begin{align} \sum_{1\leq i_1<...<i_k\leq m} \!\!\!\!\!\!\!\! \mu_{i_1}\cdots\mu_{i_k}&=\!\!\!\!\!\!\!\!\!\sum_{1\leq i_1<...<i_k\leq m} \!\!\!\!\!\!\!\! \mu_0^{i_1+...+i_k}=\!\!\!\!\!\!\!\!\!\sum_{1\leq i_1<...<i_k\leq m} \!\!\!\!\!\!\!\! \exp\left(2\pi i \frac{i_1+...+i_k}{m}\right) \end{align} At this point I need ugly combinatorics to show that this series vanishes, but it seems to me that there has to be a much simpler ansatz.

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$\mu_i$ are the solutions to $x^m-1=0$. Let $y=1-xT$. Then, $1-\mu_i T$ are solutions to $((1-y)/T)^m-1=0$ or $(1-y)^m-T^m=0$. You need the product of roots of this polynomial which is $(-1)^m\frac{1-T^m}{(-1)^m}=1-T^m$

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