Polynomial Sum Power of root

$x^{2015}+2x^{2014}+3x^{2013}+\cdots+2015x+2016=0 \cdots\cdots (1)$ $x^{2016}+2x^{2015}+3x^{2014}+\cdots +2015x^{2}+2016x=0\cdots\cdots (2)$ for $x\ne 0$ $(2)-(1)$ $x^{2016}+x^{2015}+x^{2014}+\cdots +x-2016=0$

$x^{2015}+x^{2014}+x^{2013}+\cdots +x+1=\frac{2016}{x}$ Because $x_{k}$ is root of first equation for $k=1,2,\cdots,2015$

also such that $x_{k}^{2015}+x_{k}^{2014}+x_{k}^{2013}+\cdots +x_{k}+1=\frac{2016}{x_{k}}$ or $x_{k}^{2014}+x_{k}^{2013}+\cdots +x_{k}+1=\frac{2016}{x_{k}^{2}}-\frac{1}{x_{k}}$

So $\sum_{n=0}^{2014}\, \sum_{k=1}^{2015} \left ( x_{k} \right )^{n}$ $=\sum_{k=1}^{2015}\left ( \,x_{k}^{2014}+x_{k}^{2013}+x_{k}^{2012}+\cdots +x_{k}+1 \right )$ $=\sum_{k=1}^{2015} \left ( \frac{2016} {x_{k}^{2}}-\frac{1}{x_{k}} \right ) = 2016 \sum_{k=1}^{2015} \left ( \frac{1}{x_{k}^{2}} \right ) -\sum_{k=1}^{2015} \left( \frac{1}{x_{k}} \right )$ $= 2016 \left ( \sum_{k=1}^{2015} \frac{1}{x_{k}} \right )^{2} - 4032 \sum_{1\le i\ne j\le 2015} \left ( \frac{1}{x_{i}x_{j}} \right ) -\sum_{k=1}^{2015} \left( \frac{1}{x_{k}} \right ) \cdots\cdots (3)$

Let $y_{k}=\frac{1}{x_{k}}$ for $k=1,2,\cdots,2015$

we have new equation $2016y_{k}^{2015}+2015y_{k}^{2014}+2014y_{k}^{2013}+\cdots +3y_{k}^{2}+2y_{k}+1=0$

so $(3)\, =\, 2016 \left ( \sum_{k=1}^{2015} y_{k} \right )^{2} - 4032 \sum_{1\le i\ne j\le 2015} \left ( y_{i}y_{j} \right ) -\sum_{k=1}^{2015} \left( y_{k} \right )$

from vieta formula we conclude the final answer is $2016 \left ( \frac{-2015}{2016} \right )^{2} - 4032 \left ( \frac{2014}{2016} \right ) - \left( \frac{-2015}{2016} \right )$ $= \frac{2015^{2}-2.2014.2016+2015}{2016} = \boxed{-2013}$