Rational complex exponent and proof that a matrix is unitary I need to prove that a square matrix $A$ with elements $a_{jk}$ has orthogonal rows and columns: $$a_{jk} = exp\Big(\frac{2\pi i}{n}jk\Big)_{j,k=0}^{n-1}$$
It's possible to prove it by showing that $A$ multiplied by its conjugate transpose $B=A^*$ is an identity matrix $I$ times some number $\alpha$:
$$AA^*=AB=\alpha I  = C$$
Elements of the matrix $C$ can be computed as:
$$C_{jk}=(AB)_{jk} = \sum_{m=0}^{n-1}A_{jm}B_{mk} = \sum_{m=0}^{n-1}exp\Big(\frac{2\pi i}{n}jm\Big)exp\Big(-\frac{2\pi i}{n}mk\Big)$$
If $j=k$ (diagonal elements) then $C_{jk}=\sum_{m=0}^{n-1}exp(0)=n$, however I can't figure out how to calculate $C_{jk}$ when $j \neq k$ because rules of exponentiation are somewhat different for complex exponents than for real ones.
I think the solution is related to the identity that $n$-th roots of unity, for $n > 1$, add up to $0$: ${\displaystyle \sum _{k=0}^{n-1}e^{2\pi i{\frac {k}{n}}}=0.}$
QUESTION:
How to prove  $C_{jk}=0$ for $j \neq k$ and what exponentiation rules are still applicable for rational complex exponents?
 A: Observe
\begin{align}
\sum^{n-1}_{m=0} \exp\left(\frac{2\pi i}{n} jm \right)\exp\left(-\frac{2\pi i}{n} mk \right) = \sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{n} (j-k)m \right).
\end{align}
Assume $j \neq k$ and WLOG assume $j>k$ which means
\begin{align}
(j-k) m = l m  = nq+r_m
\end{align}
where $0\leq r_m < n$. Thus, it follows
\begin{align}
\sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{n} (j-k)m \right) = \sum^{n-1}_{m=0} \exp\left(2\pi i q+ \frac{2\pi i}{n}r_m \right)= \sum^{n-1}_{m=0} \exp\left( \frac{2\pi i}{n}r_m\right). 
\end{align}
Case 1 $l$ is coprime to $n$
Then it's not hard to see that $r_m$ will take on very value $0, 1, \ldots , n-1$ which mean we could rewrite the sum as
\begin{align}
\sum^{n-1}_{k=0}\exp\left( \frac{2\pi i}{n}k\right)=0.
\end{align} 
Case 2 $l$ is not coprime to $n$.
In this case, we have that $l = dk$ and $n=dp$ where $d =\gcd (n, l)$ and $\gcd(k, p) =1$. Hence 
\begin{align}
\sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{n} lm \right)  = \sum^{n-1}_{m=0}\exp\left(\frac{2\pi i}{p}km \right)= d \sum^{p-1}_{m=0}\exp\left(\frac{2\pi i}{p}km \right)=0. 
\end{align}
A: If $0\le j,m\lt n$ and $j\ne m$, then $e^{2\pi i(j-m)}=1$ and $e^{2\pi i(j-m)\frac 1n}\ne1$. Therefore, using the formula for the sum of a geometric series,
$$
\begin{align}
\left(A\overline{A}^T\right)_{j,m}
&=\sum_{k=0}^{n-1}e^{2\pi i\frac{jk}n}e^{-2\pi i\frac{km}n}\\
&=\sum_{k=0}^{n-1}e^{2\pi i(j-m)\frac kn}\\
&=\frac{e^{2\pi i(j-m)}-1}{e^{2\pi i(j-m)\frac 1n}-1}\\[9pt]
&=0\tag{1}
\end{align}
$$
Whereas, if $j=m$, then $e^{2\pi i(j-m)\frac kn}=1$. Therefore,
$$
\begin{align}
\left(A\overline{A}^T\right)_{j,m}
&=\sum_{k=0}^{n-1}e^{2\pi i\frac{jk}n}e^{-2\pi i\frac{km}n}\\
&=\sum_{k=0}^{n-1}e^{2\pi i(j-m)\frac kn}\\[9pt]
&=n\tag{2}
\end{align}
$$
Therefore,
$$
A\overline{A}^T=nI\tag{3}
$$
