Proof of complex conjugate symmetry property of DFT According to the Proof :


  \begin{align}
X_n &amp;= \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k n}{N}}\\
X_{N-n} &amp;= \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k (N-n)}{N}}\\
&amp;=\sum_{k=0}^{N-1}x_k e^{-j 2\pi k}e^{j\frac{2\pi k n}{N}}
\end{align}
  
  Using $\exp(-j2\pi k) = 1 \quad \forall \ k$ 
  
  \begin{align}
X_{N-n} &amp;= \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}}\\
\end{align}


How is $\exp(-j2\pi k) = 1 \quad \forall \ k$ true ? Doesn't it mean $-j2\pi k = 0$ ? But that's not possible right?
Hint:

According to Euler's formula we have
$$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=\ldots$$