First of all, we know that $k ! \equiv 0(\bmod 9)$ for all $k \geq 6$. Thus, we only need to find $(1 !+2 !+3 !+4 !+5 !)(\bmod 9)$
$$
\begin{gathered}
1 ! \equiv 1(\bmod 9) \\
2 ! \equiv 2(\bmod 9) \\
3 ! \equiv 6(\bmod 9) \\
4 ! \equiv 24 \equiv 6(\bmod 9) \\
5 ! \equiv 5 \cdot 6 \equiv 30 \equiv 3(\bmod 9)
\end{gathered}
$$
Thus,
$$
(1 !+2 !+3 !+4 !+5 !) \equiv 1+2+6+6+3 \equiv 18 \equiv 0(\bmod 9)
$$
So, the remainder is $0.$
