While studying Discrete-time Fourier Transform. I found that $X[\omega]$ is periodic with $2\pi$. So when did the proof: $$ X[\omega + 2\pi] = \sum_ {n= +\infty}^{-\infty}x[n]e^{-j \omega n}e^{-j 2\pi n}$$
In the books it was written $e^{-j 2\pi n}$ =1
But I was confused that how it will be equal to 1, I tried to use Euler Equation on that but it's not possible to use that.
Anyone help me to tell how it will be equal to 1?
$e^{i\theta} = \cos(\theta) + i \sin(\theta)$. With $\theta = 2\pi \cdot k$ for some integer k, $\sin(\theta)$ will always vanish, and $\cos(\theta)$ will equal 1. In your case, $k = -n\cdot j$. Since both $n$ and $j$ are integers, their product is an integer.