In preparing some practice problems for my complex analysis students, I stumbled across the following. It is not hard to show, using Liouville's theorem, that $$\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^\infty\left(\frac{1}{z+n}+\frac{1}{z-n}\right),$$ which implies that $$-\frac{\pi z}{2}\cot(\pi z)=-\frac{1}{2}+\sum_{k=1}^\infty\zeta(2k)z^{2k},\qquad 0<|z|<1.$$ This formula predicts correctly that $\zeta(0)=-\frac{1}{2}$, and allows to calculate $\zeta(2k)$ as a rational multiple of $\pi^{2k}$ as well (in terms of Bernoulli numbers).

Is there some simple explanation why the above prediction $\zeta(0)=-\frac{1}{2}$ is valid? Perhaps there is a not so simple but still transparent explanation via Eisenstein series.

Added. Just to clarify what I mean by "simple explanation". The second identity above follows directly from the first identity, i.e. from basic principles of complex analysis: $$-\frac{\pi z}{2}\cot(\pi z)=-\frac{1}{2}+\sum_{n=1}^\infty\frac{z^2}{n^2-z^2}=-\frac{1}{2}+\sum_{n=1}^\infty\sum_{k=1}^\infty\left(\frac{z^2}{n^2}\right)^k =-\frac{1}{2}+\sum_{k=1}^\infty\zeta(2k)z^{2k}.$$ I would like to see a similar argument, perhaps somewhat more elaborate, that explains why the constant term here happens to be $\zeta(0)$, which seems natural in the light of the other terms.