Question on an identity the representation of the Riemann Zeta function using Bernoulli numbers I'm reading Serre's A Course in Arithmetic. 
In Section 4.1, we are showing that the Riemann Zeta function has a representation using the Bernoulli numbers. 
The trick was to take the logarithmic derivative of 
$$ \sin(z) = z \prod\limits_{n = 1}^{\infty} \left(1 - \frac{z^2}{n^2 \pi^2} \right). $$
I did this and got $z \cot z = 1 + 2 \sum\limits_{n = 1}^{\infty} \frac{z^2}{z^2 - n^2 \pi^2}$, as written in Serre. 
But now Serre claims that
$$ z \cot z = 1 + 2 \sum_{n = 1}^{\infty} \frac{z^2}{z^2 - n^2 \pi^2} = 1 - 2 \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{z^{2k}}{n^{2k} \pi^{2k}}. $$
How did we get the last equality? 
I thought there was some geometric series tricks going on but I failed to discover the trick after a while of hard trying. Thanks for any help! 

Note that and using some of the below, you should be ale to recover the steps in the last line above.
\begin{align}
1+2z^2\sum_{n=1}^{\infty} \frac{1}{z^2-n^2}&= 1-2z^2 \sum_{n=1}^{\infty}\frac{1}{n^2}\frac{1}{1-z^2/n^2} \\
&=1-2z^2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=0}^{\infty}\left(\frac{z}{n}\right)^{2k} \\
&= 1-2 \sum_{n=1}^{\infty}z^{2k+2}\sum_{n=1}^{\infty}\frac{1}{n^{2k+2}} \\
&=1-2 \sum_{\text{even}\, k \geq 2}\zeta(k)z^k
\end{align}
