A curious identity

$\prod_{k = 1}^{2014} \left(w^{k}-w^{-k}\right) = \displaystyle \frac{\prod_{k = 1}^{2014} \left(w^{2k}-1\right)}{w^{\sum_{k = 1}^{2014} k}} = \frac{\prod_{k = 1}^{2014} \left(e^{\frac{2k\pi i}{2015}}-1\right)}{ e^{1007 \pi i}}$

Now consider the set $\{e^{\frac{2k\pi i}{2015}}: k = 0,1,2,....,2014\}$ is the set of all solutions of $z^{2015} = 1$ Hence

$z^{2015}-1 = (z-1) \prod_{k=1}^{2014} \left(z- e^{\frac{2k\pi i}{2015}}\right)$

$1+z+z^2+....z^{2014} = \prod_{k = 1}^{2014}\left(z - e^{\frac{2k\pi i}{2015}} \right)$

Putting $z = 1$ both sides gives $\prod_{k=1}^{2014} \left(e^{\frac{2k\pi i}{2015}}-1\right) = 2015$

Hence answer is $\boxed{-2015}$