struggling to understand why Fourier basis is orthogonal Studying DSP on my own time on Coursera.

Was given a proof to why the Fourier basis is orthogonal, but I can't figure it out. Here is how it is proof goes.

Consider the Fourier basis 
$$
\left\lbrace \mathbf w^{(k)} \right\rbrace_{k=0,...,N−1}
$$

defined as:

$$
\mathbf w^{(k)}_n = e^{−j\frac{2\pi}{N} nk}
$$

Let us compute the inner product, that is

\begin{align}
&lt;\mathbf w^{(k)}, \mathbf w^{(h)} &gt; &amp;= \sum_{n=0}^{N-1} \mathbf w^{*(k)}[n] \mathbf w^{(h)}[n]\\
&amp;= \sum_{n=0}^{N-1} e^{j\frac{2\pi}{N} nk} e^{−j\frac{2\pi}{N} nh}\\
&amp; = \sum_{n=0}^{N-1} e^{-j\frac{2\pi}{N} n (h-k)}\\
&amp;= \begin{cases} N &amp; \text{if}\quad k=h\\[2ex] 0 &amp; \text{otherwise} \end{cases}
\end{align}

I do understand all of the derivation except the last step. How does one arrive at the conclusion that when $k\neq h$ the inner product is $0$. Would appreciate any help with that.
Just use the formula for the geometric series (I use $l=h-k\neq mN$):

$$\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}nl}=\frac{1-e^{-j\frac{2\pi}{N}Nl}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-e^{-j2\pi l}}{1-e^{-j\frac{2\pi}{N}l}}=\frac{1-1}{1-e^{-j\frac{2\pi}{N}l}}=0,\quad l\neq mN$$