TITLE: Is $x!-(x-1)!-(x-2)!-...-1!$ always divisible by three?
QUESTION [10 upvotes]: Just an observation:
$$x!-(x-1)!-(x-2)!-...-1!=y$$
It is observed that $y$ is always divisible by $3$. Why is this so?

REPLY [2 votes]: $$
\begin{align}
3+\overbrace{\sum_{k=3}^{n-1}(k-1)k!}^{\substack{\text{divisible by $3$}\\\text{for $n\ge3$}}}
&=1+\sum_{k=1}^{n-1}(k-1)k!\\
&=1+\sum_{k=1}^{n-1}[(k+1)!-2k!]\\
&=1+\sum_{k=2}^nk!-2\sum_{k=1}^{n-1}k!\\
&=n!-1-\sum_{k=2}^{n-1}k!\\
&=n!-\sum_{k=1}^{n-1}k!\\
\end{align}
$$