Is it even possible?

Let g(x)=x(x-1)(x-2)....(x-n). Thus f(x) can be assumed as $\sum_{k=0}^n$ $a_{k}$ $2^{k}$ g(x)/(x-k). Now we can choose the values of f(k) with the help of companion function g(x). Now obviously we will choose $a_{k}$ in a way that f(k)= $2^{k}$ . It's not difficult to see that $a_{k}$ = $(-1)^{k+1}$ 2015Ck/2015! (where nCr is the coefficient of $x^{r}$ in expansion of $(1+x)^{n}$ ). Now put x=n+1 in f(x), it becomes $\sum_{k=0}^n$ $(-1)^{k+1}$ (n+1)Ck $2^{k}$ which simplifies to $2^{n+1}$ -1. Put n=2015, and you have f(2016)= $2^{2016}$ -1= $8^{672}$ -1=(8-1)*( $8^{671}$ + $8^{670}$ +...+8²+8+1), clearly a multiple of 7.

This is my first explanation using latex :)