Find the remainder when the sum is divided by $1000$ 
Find $S \pmod{1000}$ given: $$S = \sum_{n=0}^{2015} n! + n^3 - n^2 + n - 1$$

$$S_0 = 0! + 0  - 0 + 0 -1 = 0$$
$$S_1 = 1! + 1 - 1 + 1 - 1 = 1$$
$$S_2 = 2! + 8 - 4 + 2 - 1 = 7$$
This isn't helping, so:
$n! = n(n-1)(n-2)...(1)$ but that is too complicated. 
The $n!$ term is the hardest one. Besides that:
$$S = \sum_{n=0}^{2015} n! + \frac{2015^2 2016^2}{4} - \frac{(2015)(2016)(4031)}{6} + \frac{2015(2016)}{2} - 2015$$
$$\equiv \sum_{n=0}^{2015} n! + 25(4) - 15(336)(31) + 15(8) - 15 \pmod{1000}$$
Please offer hints, thank you!

HINT : 
Since $1000=2^3\cdot 5^3$, note that
$$15!\equiv 0\pmod{1000}.$$
Hence, we have
$$\sum_{n=0}^{2015}n!\equiv \sum_{n=0}^{\color{red}{14}}n!\pmod{1000}$$
