I would like to prove that $$\binom{n}{k}=\frac{1}{2i\pi}\int_{C}\frac{(1+z)^{n}}{z^{k+1}}dz.$$
C is the circle at $0$ with radius $r>0$. I cannot get that expression, if I write the integral as $$ \int_{0}^{2\pi}\frac{(1+re^{it})^n}{r^{k+1}e^{it(k+1)}}rie^{it}dt.$$
Now using the binomial theorem I can write $(1+re^{it})^n=\sum_{k=0}^n \binom{n}{k}r^ke^{ikt}$, so I have the integral equal to $$\sum_{k=0}^n\binom{n}{k}i2\pi.$$ I think I misunderstand something because $\binom{n}{k}$ is the coefficient of $z^k$ in $(1+z)^n$, right ?.
(i) $$ (1+z)^n = \sum_{j=0}^n {n\choose j} z^j $$
(ii) $$ \int_C z^p ~ dz = 0 \text{ if }p\not = -1, 2\pi i \text{ if }p=-1 $$