Will you divide it?

We note that $n!$ is divisible by 14 for $n \ge 7$ . Therefore, we have:

$\begin{aligned} \sum_{n=1}^{200} n! & \equiv \sum_{n=1}^6 n! \text{ (mod 14)} \\ & \equiv (1! + 2! + 3! + 4! + \color{#3D99F6}{5! + 6!}) \text{ (mod 14)} & \small \color{#3D99F6}{\text{Note that } 5!+6! = 7 \cdot 5! \text{ which is divisible by }7} \\ & \equiv (1! + 2! + 3! + 4! + \color{#3D99F6}{0}) \text{ (mod 14)} \\ & \equiv (1 + 2 + 6 + 24) \text{ (mod 14)} \\ & \equiv (1 + 2 + 6 + 10) \text{ (mod 14)} \\ & \equiv 19 \text{ (mod 14)} \\ & \equiv \boxed{5} \text{ (mod 14)} \end{aligned}$

Hey this is really easy

$1! =1$

$2! =2$

$3! =6$

$4!=24$

But in and after $5!$ the values are going to end with $0$ due to 2 × 5

So it comes down to $(1!+2!+3!+4!)(mod 14)$ = $33(mod14)$ = $\boxed{5}$