10. For a positive integer $n$, define $n!=n(n-1)(n-2) \cdots 2 \cdot 1$, and let $S_{n}=n!\left[\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}-1\right]$, then $S_{2017}=$ $\qquad$ .
$\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$, so $S_{n}=n!\left[-\frac{1}{(n+1)!}\right]=-\frac{1}{n+1}$. Therefore, $S_{2017}=-\frac{1}{2018}$.