10. Let $S_{n}=n!\left(\sum_{k=1}^{n} \frac{k}{(k+1)!}-1\right)$. Then $S_{2017}=$ $\qquad$ .
10. $-\frac{1}{2018}$.

From $\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$, we get
$$
\begin{array}{l}
S_{n}=n!\left(\sum_{k=1}^{n}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)-1\right) \\
=n!\left(-\frac{1}{(n+1)!}\right)=-\frac{1}{n+1} .
\end{array}
$$

Therefore, $S_{2017}=-\frac{1}{2018}$.