How to prove that:
Note that $\displaystyle \sum\limits_{k = 1}^5 {k!} = 153$ is a multiple of nine.
Also, if $\displaystyle N\ge 6$ then $\displaystyle 6!={\color{blue}6}\cdot 5\cdot 4\cdot{\color{blue}3}\cdot 2={\color{blue}720}$ which is a multiple of nine.
If $\displaystyle N\ge 6$ then $\displaystyle N!$ is a multiple of nine. WHY?