Just Observe The Terms

All numbers greater than $3!$ are divisible by $12$ . So remainder is $1!+2!+3!=1+2+6=\boxed9$ .

$\text{We know } n! = (n)\times(n-1)\times(n-2)\times ...\times (2)\times(1)\\ \text{That means, for }n \geq m, n! \text{ is divisible by m, as n! contains m in it.}\\ \text{For example, 3! = }3\times2\times1 = 6, \text{ if m = 2, then 3! is divisible by 2, 3!/2 = 6/2 = 3} \\ \text{Using the above fact, we now gonna apply that in here. Taking n as 1 to 2016, and m as 12. All the factorials } \geq 12! \\\text{is divisible by 12 and leaves the remainder "0".}\\ \text{Therefore the remainder of 1! + 2! + ... + 2016! is when Sum-of n! divided by 12 for all n } \leq \text{ 11}\\ => 1! + 2! + 3! + 4! + ... + 11! = 43954713 \\ =>1! + 2! + 3! + 4! + ... + 11! = 12 * 3662892 + 9$

Therefore the remainder is 9.