From what I understand, the main "point" of the Laurent series is that we should be able to derive it easily (e.g. by stitching together known Taylor series), and then exploit its uniqueness to say that these coefficients are the same as $c_n=\frac{1}{2\pi i}\oint_\circ \frac{f(z)}{(z-a)^{n+1}} dz$, from which we can calculate e.g. the residue $2\pi c_{-1}$.
But to actually do this, we need one crucial fact: the series $\sum_{n\in\mathbb{Z}}c_n(z-a)^n$ given by $c_n=\frac{1}{2\pi i}\oint_\circ \frac{f(z)}{(z-a)^{n+1}} dz$ is actually a valid Laurent series for $f(z)$, i.e. where it converges, it converges to $f(z)$.
How does one prove this? It feels like it should follow easily from Cauchy's integral formula, but my brain doesn't seem to be working sensibly at the moment.