Let $f(z)=\sum_{n=0}^{\infty}c_nz^n$ have radius of convergence $R>0.$
Use the fact that $$\sum\limits_{n=0}^{\infty}\int_\gamma c_n z^ndz=\int_\gamma \sum\limits_{n=0}^{\infty}c_nz^ndz=\int_\gamma f(z)dz$$ to show that $$c_k=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z^{k+1}}dz \qquad(0<r<R)$$ where $\gamma$ is the circle of radius $r$ centred at the origin, traversed once in the positive direction.
From the information given, I could construct a parametrisation $\gamma(t)=re^{it}$, then $\gamma '(t)=rie^{it}$. Using the identities gives, I could isolate the $c_k$ term by moving all other terms to the other side of the equality. Since $f(z)$ is in the final expression, I considered starting with $$\int_\gamma c_kz^kdz=\int_\gamma f(z)dz-\int_\gamma c_0dz-\cdots-\int_\gamma c_{k-1}z^{k-1}dz-\int_\gamma c_{k+1}z^{k+1}dz-\cdots-\int_\gamma c_nz^ndz$$ And I thought maybe the $\frac{1}{2\pi i}$ term came from using the parametrisation with $t\in[0,2\pi]$.
But I'm not sure this is a good idea. I end up with a LHS looking like: $$c_k\left(\frac{r^2\left(e^{2i\pi (k+1)}-1\right)}{k+1}\right)$$ It's a little hopeful since the $z^{k+1}$ term could be of the form $re^{i(k+1)}$, but over all looking too messy. Especially since I can't think of a way to simplify the RHS either.
What is the best way to approach this?
You have
$$ \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z^{k+1}} dz=\frac{1}{2\pi i} \int_{\gamma} \frac{\sum_{n=0}^{\infty} c_nz^n}{z^{k+1}} = \frac{1}{2\pi i} \sum_{n=0}^{\infty} \int_{\gamma} \frac{c_nz^n}{z^{k+1}}=\frac{1}{2\pi i} \sum_{n=0}^{\infty}c_n \int_{\gamma} z^{n-k-1} $$
and now refer to the result that
$$ \int_{\gamma} z^m dz= \begin{cases} 2 \pi i & \text{if} \; m=-1\\ 0 & \text{otherwise} \end{cases} $$
So
$$ \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z^{k+1}} dz=\frac{1}{2\pi i} \sum_{n=0}^{\infty}c_n \int_{\gamma} z^{n-k-1} = \frac{1}{2\pi i}\cdot 2\pi i\cdot c_k=c_k $$
as the integrals vanish unless $n=k$