A triangle polynomial

Built $p(x)$ with polynomial Lagrange:

$\begin{aligned} p(x)=&\frac{(x-1)(x-2)\dots(x-2014)}{2014!\cdot0!}+\frac{x(x-2)\dots(x-2014)}{2013!\cdot1!}+\dots+ \\ &+\frac{x(x-1)\dots(x-2012)(x-2014)}{1!\cdot2013!}+\frac{x(x-1)\dots(x-2013)}{0!\cdot2014!} \end{aligned}$

where the l' $i-$ th term is built such as $p(i)=(-1)^i$ $\forall i\in{0,1,\dots,2014}$ (substituted).

So evaluate $p(2015)$ :

\begin{equation*} p(2015)=1+\frac{2015!}{2014!\cdot1!}+\frac{2015!}{2013!\cdot2!}+\dots+\frac{2015!}{2!\cdot2013!}+\frac{2015!}{1!\cdot2014!}. \end{equation*}

Follows that $p(2015)$ can be rewritten:

\begin{equation*} p(2015)=\binom{2015}{0}+\binom{2015}{1}+\dots+\binom{2015}{2013}+\binom{2015}{2014}=2^{2015}-1. \end{equation*}

Now you just have to find the last $4$ digits of $2^{2015}-1$ but: \begin{equation*} 2^{2015}\equiv{2^{15}}\pmod{625} \text{ because } \phi(625)=500 \end{equation*} so \begin{equation*} 2^{2015}\equiv{268}\pmod{625}. \end{equation*} Consequently the last $4$ digits of $2^{2015}$ can be $268+625k$ but it is true that the last $4$ digits of a power of $2$ must be divisible at least by $2^4$ and this is true only if $k=4$ ;

$2^{2015}-1\equiv{\fbox{2767}}\pmod{10000}.$

Using method of differences, we get the array

1 -1 1 -1 1 -1........ -1 1 -2 2 -2 2 -2 ......... 2 4 -4 4 -4 .........4 -8 8 -8 8.......8 ................. 2^2014. Because of the polynomial degrees, this 2^2014 must remain constant. Hence f(2015)=22^2014+2^2013.....+2+1=2^2015-1. Taking mod 10000 we get 2767.