Let's have a try

$\begin{aligned} S_n & = n! - n!\left(\frac 1{2!} + \frac 2{3!} + \frac 3{4!} + \cdots + \frac n{(n+1)!} \right) \\ & = n! - n!\sum_{k=1}^n \frac k{(k+1)!} \\ & = n! - n!\sum_{k=1}^n \frac {k+1-1}{(k+1)!} \\ & = n! - n!\sum_{k=1}^n \left(\frac 1{k!} - \frac 1{(k+1)!}\right) \\ & = n! - n! \left(\frac 11 - \frac 1{(n+1)!}\right) \\ & = n! - n! + \frac 1{n+1} \\ & = \frac 1{n+1} \end{aligned}$

Therefore, $S_{2017} = \dfrac 1{2018}$ , $\implies a = \boxed{1009}$ .

Note that $\displaystyle\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$ ,so $\displaystyle\sum_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}\Longrightarrow\ S_n =\frac{1}{n+1}\Longrightarrow\ S_{2017} =\frac{1}{2018}$ ,so answer is $\boxed{1009}$