Integral related to binomial coefficient $\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{(1+e^{ix})^n}{e^{ixk}}~dx$

Question

I want to show that $$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{(1+e^{ix})^n}{e^{ixk}}~dx={n\choose k}$$ My attempts were very poor, since I'm not an expert in contour integration.I thought that, in order to have factorials on the right, the integral could be reduced to a Beta function, but substituting $t=e^{ix}$ doesn't work because of the bounds $-\pi,\pi$. I am wondering if anyone could give me a shove in the right direction.

Answer 1

$$I=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{(1+e^{ix})^n}{e^{ixk}}\text{d}x$$

Let $z=e^{ix}$ $$I=\frac{1}{2\pi}\oint\frac{(1+z)^n}{z^k}\frac{1}{iz}dz=\frac{1}{2\pi i}\oint\frac{(1+z)^n}{z^{k+1}}dz=\frac{1}{2\pi i}\cdot 2\pi i \cdot Res(z=0)$$

where the contour is unit circle, counter-clockwisely.

$$Res(z=0)=\binom{n}{k}$$

therefore,

$$I=\binom{n}{k}$$

Answer 2

Hints:

$$\left(1 + e^{\mathrm ix}\right)^n =\sum_{\ell = 0}^{n} \binom{n}{\ell} e^{\mathrm i\ell x}$$

and

$$\int_{-\pi}^{\pi}e^{\mathrm imx}\mathrm d x = \begin{cases}2\pi & \text{if $m=0$}\\ 0 &\text{otherwise}\end{cases}$$