What I want to prove: Suppose $\lambda \in (-\pi,\pi]$ are natural frequencies at time $n$. Then for every $\lambda_j$ define a vector $e_j^n = \frac{1}{\sqrt{n}} \left(e^{i \lambda_j},e^{2i\lambda_j},\ldots,e^{ni\lambda_j}\right)^T$. The $n$ vectors form an orthonormal basis for $\mathbb{C}^n$.
$n$ points are called natural frequencies at time $n$: $x_1,\ldots,x_n = \ldots,-\frac{4 \pi}{n}, - \frac{2 \pi}{n},0, \frac{2\pi}{n},\frac{4 \pi}{n},\ldots \subset (-\pi,\pi]$. (The value 0 is always a natural frequency; for odd $n$ there are $(n−1)/2$ positive and negative natural frequencies, situated symmetrically about 0; for even $n$ the value $\pi$ is a natural frequency, and the remaining natural frequencies are $n/2 − 1$ points in $(0, \pi)$ and their reflections.)
My attempt: Suppose $j \neq k$ then \begin{align*} \langle e_j^n,e_k^n \rangle &= \frac1n \langle e^{i \lambda_j},e^{2i\lambda_j},\ldots, e^{ni\lambda_j} ; e^{i \lambda_k},e^{2i\lambda_k},\ldots, e^{ni\lambda_k} \rangle \\ &= \frac1n \left( e^{i(\lambda_j + \lambda_k)}+\cdots + e^{ni(\lambda_j + \lambda_k)}\right). \end{align*} How is this zero? And how do is the length of each $e^n_j$ one as $$ ||e^n_j|| = \langle e^n_j ; e^n_j \rangle = \frac1n \left( e^{2i \lambda_j} + \cdots +e^{2i n \lambda_j} \right).$$ And lastly how do these vectors span $\mathbb{C}^n$? Thanks for any assistance.
Hint: note that for any integer $p$, $$ e^{p\,i \lambda_j - p\,i\lambda_k} = \left( e^{i(\lambda_j - \lambda_k)}\right)^p $$ In order to compute the inner products, you can use the formula $$ \sum_{p=0}^{n-1} \xi^p = \frac{\xi^{n} - 1}{\xi - 1}, \quad \xi \neq 0 $$