Let $f_1(x)=(x+53)(x+106)(x+159)\cdots(x+2014)$. Define $f_2(x)=f_1(x+1)-f_1(x)$ ; $f_3(x)=f_2(x+1)-f_2(x)$ ; $\ldots$ ; $f_n(x)=f_{n-1}(x+1)-f_{n-1}(x)$. What is the smallest value of $n$ for which $f_n(x)$ is a constant polynomial?
The only thing I could figure out was that 106= 2(53); 159 = 3(53) and 2014=38(53). That's about it, any suggestions?