Divisible by this year??? v2015 (Part 2: A Jinx from the past!!!)

Let $f(n) = (n+2)! \times (n + 0)! \times (n + 1)! \times (n + 4)!$

The domain of this function is $[0, + \infty )$

So let's first check $f(0)$ .

$(n+2)! \times (n + 0)! \times (n + 1)! \times (n + 4)!$ $= (0+2)! \times (0 + 0)! \times (0 + 1)! \times (0 + 4)!$ $= 2! \times 0! \times 1! \times 4!$ $= 48$

There are $0$ trailing zeroes in $48$ . And since $0$ is divisible by $2015$ , the answer is 0