Think Simple

$\begin{aligned} S & = \sum_{\color{#3D99F6}{k}=1}^{2015} \left(\frac{1}{k!} - \frac{k^2}{(k+1)!} \right) \quad \quad \small \color{#3D99F6}{\text{Sorry, I am used to using } k \text{ and reserving } w \text{ for roots of unity.}} \\ & = \sum_{k=1}^{2015} \left(\frac{1}{k!} - \frac{k^2-1+1}{(k+1)!} \right) \\ & = \sum_{k=1}^{2015} \left(\frac{1}{k!} - \frac{(k+1)(k-1)+1}{(k+1)!} \right) \\ & = \sum_{k=1}^{2015} \left(\frac{1}{k!} - \frac{k-1}{k!} - \frac{1}{(k+1)!} \right) \\ & = \sum_{k=1}^{2015} \left(\frac{1}{k!} - \frac{1}{(k-1)!} + \frac{1}{k!} - \frac{1}{(k+1)!} \right) \\ & = \sum_{k=1}^{2015} \left(- \frac{1}{(k-1)!} + \frac{1}{k!} + \frac{1}{k!} - \frac{1}{(k+1)!} \right) \\ & = - \sum_{k=\color{#3D99F6}{0}}^{\color{#3D99F6}{2014}} \frac{1}{k!} + \sum_{k=1}^{2015} \frac{1}{k!} + \sum_{k=1}^{2015} \frac{1}{k!} - \sum_{k=\color{#D61F06}{2}}^{\color{#D61F06}{2016}} \frac{1}{k!} \\ & = - \frac{1}{0!} + \frac{1}{2015!} + \frac{1}{1!} - \frac{1}{2016!} = \frac{1}{2015!}- \frac{1}{2016!} = \frac{2015}{2016!} \\ & \\ \Rightarrow 2016!S & = \boxed{2015} \end{aligned}$