Show that if $f$ is holomorphic in an annulus $V={r<|z-c|<R}$ then the integral
$$\int_{|z-a|=\rho}fdz$$ has the same value for all $\rho$, $r<\rho<R$.
My approach: Suppose a function $f(z)$ holomorphic on the annulus $r <|z −c| < R$ has two Laurent series:
$$f(z)=\sum_{n=-\infty}^{\infty}a_{n}\left(z-c\right)^{n}=\sum_{n=-\infty}^{\infty}b_{n}\left(z-c\right)^{n}$$
Multiply both sides with $\left(z-c\right)^{-k-1}$, where k is an arbitrary integer, and integrate on a path γ inside the annulus,
$$\oint_{\gamma}\sum_{n=-\infty}^{\infty}a_{n}\left(z-c\right)^{n-k-1}\mathrm{d}z=\oint_{\gamma}\sum_{n=-\infty}^{\infty}b_{n}\left(z-c\right)^{n-k-1}\mathrm{d}z$$
The series converges uniformly on $\epsilon\leq|z-c|\leq R-\epsilon$, where ε is a positive number small enough for γ to be contained in the constricted closed annulus, so the integration and summation can be interchanged. Substituting the identity
$$\oint_{\gamma}(z-c)^{n-k-1}dz=2\pi i\delta_{nk}$$
into the summation yields
$$a_k=b_k$$
Hence the Laurent series is unique.
Can anyone assist help explain why $$\oint_{\gamma}(z-c)^{n-k-1}dz=2\pi i\delta_{nk}$$ is the case to me?
Hint: consider a contour that looks like this: