2017 here I come!

Since $P(1)= P(2) = P(3) = ... = P(2016) =0$ , we can see that ${1,2,3,...,2016}$ are the roots of $P(x)$ . Then, the largest root of $P(x)$ is $2016$ .

A polynomial of degree $2017$ which has the numbers from $1$ to $2016$ as roots cam be $P(x)=(x-1)(x-2)(x-3)\cdots(x-2016)(x-a)$ $\Rightarrow P(0)=-a\cdot 2016! =2018! \Rightarrow a=-2017\cdot 2018$ Then the other root must be negative, so the greatest root is $2016$

It's given that P(x) is a polynomial of degree 2008, u also know 20016 roots of P(x) ....(which are 1,2,3.....2016) let the other root be m P(x) can be written as (x-1)(x-2)(x-3)(x-4)......(x-2016)(x-m ) .....now Use P(0) = 2009! when we plug P(0) we get the value 2017*2018=4070306