Almost halfway through 2017!

Relevant wiki: Telescoping Series - Product

$\displaystyle 1 - \frac{1}{k^2} = \frac{k^2 - 1}{k^2} = \frac{(k - 1)(k + 1)}{k^2}$

Thus,

$\displaystyle \Bigg(1 - \frac{1}{2^2}\Bigg)\Bigg(1 - \frac{1}{3^2}\Bigg) \cdots \Bigg(1 - \frac{1}{2017^2}\Bigg) = \prod_{k \ = \ 2}^{2017} \frac{(k - 1)(k + 1)}{k^2}$

$\displaystyle \frac{1 \cdot 3}{2^2} \cdot \frac{2 \cdot 4}{3^2} \cdot \frac {3 \cdot 5}{4^2} \cdots \cdots \frac{2015 \cdot 2017}{2016^2} \cdot \frac{2016 \cdot 2018}{2017^2} = \frac{2018}{2 \cdot 2017}$