fourier transform - Proof of complex conjugate symmetry property of DFT

According to the Proof :

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

Using $\exp(-j2\pi k) = 1 \quad \forall \ k$

$$\begin{align} X_{N-n} &= \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}}\\ \end{align}$$

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?

Answer

Hint:

According to Euler's formula we have

$$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=\ldots$$