Is $x!-(x-1)!-(x-2)!-...-1!$ always divisible by three?

Just an observation:

$$x!-(x-1)!-(x-2)!-...-1!=y$$

It is observed that $y$ is always divisible by $3$. Why is this so?

This does not hold for $x < 3$. Otherwise, observe that $$ x! - \sum_{j=1}^{x-1} (x-j)! = x! - \sum_{j=1}^{x-3} (x-j)! - (2! + 1!) $$ the former is divisible by $3$ since $x_j \geq 3$ for all such $j$. The latter is divisible by $3$ with simple arithmetic. Hence the entire expression is divisible by $3$, when $x \geq 3$.

A bit of an interesting pattern that has an easy explanation. Note, that this is only true where $n > 2$, since $2! - 1! = 2 - 1 = 1$ which is evidently not divisible by 3.

This is due to a simple consequence from the fact that for every $n >2$, $n!$ is divisible by $3$. Why? $$n! = n \cdot n - 1 \cdot n-2 \ \cdot \ ....... \ \cdot 3 \cdot 2 \cdot 1$$ And $3$ is a factor...so obviously $3$ divides $n!$. Without going into modular arithmetic as it may not be an area you are familiar with, if $x, y$ are divisible by $3$, then so is $x - y$ (You can try verifying this by yourself if you don't think this is convincing).

So the only possible violations in $n! - (n-1)! - (n-2)! - .... - 1!$ being divisible by $3$ is from $-2! - 1!$ since $$\underbrace{n! - (n-1)! - (n-2)! - (n-3)! - (n-4)! - (n-5)! - ..... }_\text{All divisible by 3} - 2! - 1!$$ Noting that $2 + 1 = 3$ (subtracting by 3 at the end), then the result follows immediately.

(I just want to say that it is these exact same patterns that to me seemed extraordinary at the time, that made me get interested in mathematics).