Product expansion of $\sin x$. I am reading through some Complex Analysis notes and have come to product expansion of $\sin x$, which aim to show that $$ \frac{\pi^2}{\sin^2( \pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z-n)^2}$$
To prove this we first note that both sides of the equality have double poles at the integers. T
The Laurent expansion of the right-hand side near $n \in \mathbb{Z}$ begins 
$$\frac{1}{(z-n)^2} + \text{holomorphic}$$
I am lost as to where this expression is coming from. 
Are we looking for the Laurent expression of the series, $\sum_{n \in \mathbb{Z}} \frac{1}{(z-n)^2}$?  Why is the given expression correct?

To prove the equation, we start by looking at the leading terms (the terms in $1/(z-n)^2$ and $1/(z-n)$) of the Laurent expansions of both sides about the pole $z=n$.  Once we show that these are equal, the difference between the two sides has a removable singularity at each $z=n$, and thus extends to an entire function. 
The next step might be to show that this entire function is bounded, and therefore (by Liouville) constant.  The final step will be to show that this constant is $0$.
