$$ w_k[n] = e^{j\frac{2p}{N}nk} $$ where
- $k = 0,1,...,N-1$ is the index of the vector
- $n = 0,1,...,N-1$ is the index of the element of each vector (i.e. $N-1$ vectors each of size $N-1$) is an orthogonal basis in $\mathbb{C}^N$
In vector notation $$ w^{(k)}_{k = 0,1,...,N-1} $$ with $$ w_n^{(k)} = e^{j\frac{2\pi}{N}nk} $$ is an orthogonal basis in $\mathbb{C}^N$
Proof \begin{align*} \langle\textbf{w}^{(k)}, \textbf{w}^{(h)} \rangle &=\sum_{n = 0}^{N-1} \big(e^{j\frac{2\pi}{N}nk\big)^\star} e^{j\frac{2\pi}{N}nh}\\ &=\sum_{n = 0}^{N-1} e^{j\frac{2\pi}{N}(h-k)n}\\ &=\begin{cases} N &\text{ for } h = k\\ \frac{1-e^{j2\pi(h-k)}}{1-e^{j\frac{2\pi}{N}(h-k)}} = 0 &\text{ otherwise (i.e. for $h\ne k$)} \end{cases} \end{align*}
My question: How do we prove that $$ \sum_{n = 0}^{N-1} e^{j\frac{2\pi}{N}(h-k)n} = \frac{1-e^{j2\pi(h-k)}}{1-e^{j\frac{2\pi}{N}(h-k)}} = 0 \quad \text{for $h\ne k$} $$ ?