Consider the following contour integral:
$$\oint\limits_{C_N} f(z) \pi \cot(\pi z) dz$$
where $f(z)$ is a rational function with degree(denominator) exceeds degree(numerator) by at least $2$ and $C_N$ is a square contour centered at $0$ with sides of length $2N+1$.

We can use residues to calculate the value of the integral:
$
\begin{eqnarray}
& & \oint\limits_{C_N} f(z) \pi \cot(\pi z) dz \\
& = & 2 \pi i \left[ \sum\limits_{\text{integer} \; k} \text{Res}\left[ \pi f(z) \cot(\pi z); k\right] + \sum\limits_{\text{poles} \; z^* \; \text{of f}} \text{Res}\left[ \pi f(z) \cot(\pi z); z^*\right] \right] \\
& = & 2 \pi i \left[ \sum\limits_{\text{integer} \; k}^{\;\;\;\;\;\;\;\;\;\;\;\; \prime} f(k) + \sum\limits_{\text{poles} \; z^* \; \text{of f}} \text{Res}\left[ \pi f(z) \cot(\pi z); z^*\right] \right]
\end{eqnarray}
$

Where $\sum\limits_{\text{integer} \; k}^{\;\;\;\;\;\;\;\;\;\;\;\; \prime} f(k)$ is taken to be the sum of $f(k)$ for all integer $k$ such that $z=k$ is not a pole of $f(z)$, since:

$
\begin{eqnarray}
& & \text{Res}\left[ \pi f(z) \cot(\pi z); k\right] \\
& = & f(z) \pi \frac{\cos(\pi z)}{\pi \cos(\pi z)} \bigg|_{z=k} \\
& = & f(k) \;\;\; \text{if } z=k \text{ not a pole of } f(z)
\end{eqnarray}
$

Now, as it turns out, as $N \rightarrow \infty$, the line integral around the contour $C_N$ goes to $0$. This is given without proof, since the proof is annoying.
Finally, we get the following result:

$
\begin{eqnarray}
0 & = & \oint\limits_{C_N} f(z) \pi \cot(\pi z) dz \\
0 & = & 2 \pi i \left[ \sum\limits_{\text{integer} \; k}^{\;\;\;\;\;\;\;\;\;\;\;\; \prime} f(k) + \sum\limits_{\text{poles} \; z^* \; \text{of f}} \text{Res}\left[ \pi f(z) \cot(\pi z); z^*\right] \right] \\
\sum\limits_{\text{integer} \; k}^{\;\;\;\;\;\;\;\;\;\;\;\; \prime} f(k) & = & -\sum\limits_{\text{poles} \; z^* \; \text{of f}} \text{Res}\left[ \pi f(z) \cot(\pi z); z^*\right]
\end{eqnarray}
$

Example: $\zeta(2)$

$
\begin{eqnarray}
\sum\limits_{k=1}^\infty \frac{1}{k^2} & = & \frac{1}{2} \left( -\text{Res}\left[ \frac{\pi}{z^2} \cot(\pi z); 0\right] \right) \\
& = & -\frac{1}{2} \left[ \frac{\pi \cos(\pi z)}{z^2 \sin(\pi z)} \right] \bigg|_{z=0} \\
& = & -\frac{1}{2 z^3} \left(1 – \frac{\pi^2 z^2}{2} + \ldots \right) \frac{1}{1 – \frac{\pi^2 z^2}{6} + \ldots} \\
& = & -\frac{1}{2 z^3} \left(1 – \frac{\pi^2 z^2}{2} + \ldots \right)\left(1 + \frac{\pi^2 z^2}{6} + \ldots \right) \\
& = & -\frac{1}{2 z^3} \left(1 – \frac{\pi^2 z^2}{3} + \ldots \right) \\
& = & -\frac{1}{2 z^3} + \frac{\pi^2}{6 z} + \ldots
\end{eqnarray}
$

Of course, only the $\frac{1}{z}$ term matters, so the answer is:

$$\zeta(2) = \frac{\pi^2}{6}$$

An alternate method to compute the even zeta values is to look at the Taylor expansion of the cotangent function around $0$.

Consider $\frac{\sin x}{x}$ which is $1$ as $x \rightarrow 0$.

$
\begin{eqnarray}
\frac{\sin x}{x} & = & 1 \cdot \left(1 – \frac{x}{\pi} \right) \left(1 – \frac{x}{-\pi} \right) \left(1 – \frac{x}{2 \pi} \right) \left(1 – \frac{x}{-2 \pi} \right) \cdots \\
& = & \left(1 – \frac{x^2}{\pi^2} \right) \left(1 – \frac{x^2}{2^2 \pi^2} \right) \cdots \\
& = & \prod\limits_{n=1}^\infty \left(1 – \frac{x^2}{n^2 \pi^2} \right) \\
\ln\left( \frac{\sin x}{x} \right) & = & \sum\limits_{k=1}^\infty \ln \left(1 – \frac{x^2}{n^2 \pi^2} \right) \\
& = & -\sum\limits_{k=1}^\infty \sum\limits_{n=1}^\infty \frac{1}{n} \left( \frac{x^2}{k^2 \pi^2}\right)^n \;\;\; \bigg| |x| < \pi \\ & = & -\sum\limits_{n=1}^\infty \frac{x^{2n}}{n \pi^{2n}} \sum\limits_{k=1}^\infty \frac{1}{k^{2n}} \\ & = & -\sum\limits_{n=1}^\infty \frac{x^{2n}}{n \pi^{2n}} \zeta(2n) \\ \frac{\partial}{\partial x} \left(\ln \sin x - \ln x \right) & = & \frac{\partial}{\partial x} \left[-\sum\limits_{n=1}^\infty \frac{x^{2n}}{n \pi^{2n}} \zeta(2n) \right] \\ \cot x - \frac{1}{x} & = & -\sum\limits_{n=1}^\infty \frac{2 x^{2n-1}}{\pi^{2n}} \zeta(2n) \\ \left( -\frac{x}{3} - \frac{x^3}{45} - \frac{2 x^5}{945} - \cdots \right) & = & -\sum\limits_{n=1}^\infty \frac{2 x^{2n-1}}{\pi^{2n}} \zeta(2n) \\ \left( x\frac{\pi^2}{6} + x^3\frac{\pi^4}{90} + x^5\frac{\pi^6}{945} + \cdots \right) & = & \sum\limits_{n=1}^\infty x^{2n-1} \zeta(2n) \\ \end{eqnarray} $ Equating the $x^k$ terms, we get: $ \begin{eqnarray} \zeta(2) & = & \frac{\pi^2}{6} \\ \zeta(4) & = & \frac{\pi^4}{90} \\ \zeta(6) & = & \frac{\pi^6}{945} \\ & \vdots & \end{eqnarray} $

Consider:

$
\begin{eqnarray}
1 + \frac{1}{3^k} + \frac{1}{5^k} + \frac{1}{7^k} + \frac{1}{9^k} + \cdots & = & \zeta(k) – \frac{1}{2^k} – \frac{1}{4^k} – \cdots \\
& = & \zeta(k) – \frac{1}{2^k} \left(1 + \frac{1}{2^k} + \frac{1}{3^k} + \cdots \right) \\
& = & \left(1 – \frac{1}{2^k} \right) \zeta(k) \\
1 + \frac{1}{5^k} + \frac{1}{7^k} + \frac{1}{11^k} + \frac{1}{13^k} + \cdots & = & \left(1 – \frac{1}{2^k} \right) \zeta(k) – \frac{1}{3^k} – \frac{1}{9^k} – \frac{1}{15^k} – \cdots \\
& = & \left(1 – \frac{1}{2^k} \right) \zeta(k) – \frac{1}{3^k}\left(1 – \frac{1}{3^k} – \frac{1}{5^k} – \cdots \right) \\
& = & \left(1 – \frac{1}{2^k} \right) \left(1 – \frac{1}{3^k} \right) \zeta(k) \\
& \vdots
\end{eqnarray}
$

What we see is that we can keep removing terms divisible by prime $p$ by multiplying $\zeta(k)$ by $\left(1 – \frac{1}{p^k}\right)$. In general,

$$\zeta(n) = \prod\limits_{\text{prime } p} \left(1 – \frac{1}{p^n} \right)^{-1}$$