What insight is supposed to be gained from this complex analysis exercise? Let $C_0$ denote the circle centered around some point $z_0\in\mathbb{C}$ with radius $R$. We can parametrize this circle like this:
$$\begin{array}{cc} z(\theta)=z_0+Re^{i\theta}, & \theta \in [-\pi,\pi]\end{array}$$
We are asked to show that:
$$\int_{C_0}(z-z_0)^{n-1}dz = \left\lbrace \begin{array}{ll} 0 & n\in\mathbb{Z}\backslash\{0\} \\ 2\pi i & n=0\end{array}\right. $$
I have completed the exercise but do not quite understand what this gives us. Could someone please provide some form of intuition for this?

What this gives is something remarkably powerful. Specifically, the result tells you that there's a way to isolate the various terms of a polynomial, and in fact a power series.
Think of it this way: take $z_0=0$. We then have that
\begin{equation*}
\frac{1}{2\pi i}\int_{C_0}{z^{n-1}}\,dz=\begin{cases}1&n=0,\\0&\text{otherwise}.\end{cases}
\end{equation*}
Any time you can obtain the Kronecker delta, it gives you the chance to select a specific piece of your input. In this case, let $P$ be a polynomial: $P(x)=a_0+a_1x+a_2x^2+\ldots+a_nx^n$.
Consider the integral
\begin{equation*}
\frac{1}{2\pi i}\int_{C_0}{\frac{P(z)}{z^{k+1}}}\,dz.
\end{equation*}
Expanding this out, we have
\begin{align*}
\frac{1}{2\pi i}\int_{C_0}{\frac{P(z)}{z^{k+1}}}\,dz
&=\frac{1}{2\pi i}\int_{C_0}{\left(a_0z^{-k-1}+a_1z^{(1-k)-1}+\cdots+a_nz^{(n-k)-1}\right)}\,dz\\
&=\frac{a_0}{2\pi i}\int_{C_0}{z^{(0-k)-1}}\,dz+\frac{a_1}{2\pi i}\int_{C_0}{z^{(1-k)-1}}\,dz+\cdots+\frac{a_n}{2\pi i}\int_{C_0}{z^{(n-k)-1}}\,dz.
\end{align*}
By the result you proved, this is extremely easy to compute. Each integral $\int_{C_0}{z^{(\ell-k)-1}}\,dz$ is $0$ unless $\ell=k$, so we get
\begin{align*}
\frac{1}{2\pi i}\int_{C_0}{\frac{P(z)}{z^{k+1}}}\,dz
&=0+\cdots+0+a_k+0+\cdots+0\\
&=a_k.
\end{align*}
Therefore, given a polynomial function $P$, we can extract the coefficients of the polynomial by integrating $P(z)$ on a circle about $0$. This works for power series as well... and changing the center of the circle just moves us from a power series in $z$ to a power series in $(z-z_0)$. (That is, from a Maclaurin series to a Taylor series.)
In other words, if $f$ is an analytic function, we can compute its derivatives at $0$ by computing integrals! We just need to compensate for the factor of $k!$ that appears in the coefficients of the power series, obtaining
\begin{equation*}
f^{(k)}(0)=\frac{k!}{2\pi i}\int_{C_0}{\frac{f(z)}{z^k}}\,dz.
\end{equation*}
In fact,
\begin{equation*}
f^{(k)}(z_0)=\frac{k!}{2\pi i}\int_{C_0}{\frac{f(z)}{(z-z_0)^k}}\,dz,
\end{equation*}
if the circle is centered at $z_0$.
P.S.: It's actually sufficient for the circle to contain $z_0$, even if it has a different center. In fact, it doesn't even need to be circular! Explaining that leads to some truly beautiful mathematics... all the way to Cauchy's integral formula, and its many fascinating consequences.
