Math mix: series sum + factorial

We can show $\sum_{x=1}^{n}(x!x)=(n+1)!-1$ by induction on $n$ , with the case $n=1$ being trivial: $\sum_{x=1}^{n+1}(x!x)=(n+1)!-1+(n+1)!(n+1)=(n+2)!-1$

Relative wiki : Telescoping Series - Sum

We have:

$\displaystyle \begin{aligned} \sum_{x=1}^n x!\cdot x&=\sum_{x=1}^n \left(x!\cdot (x+1-1)\right)\\ &=\sum_{x=1}^n \left(x!\cdot (x+1)\right)- \sum_{x=1}^n x!\\ &=\sum_{x=1}^n (x+1)!-\sum_{x=1}^n x!\\ &=(n+1)!-1 \end{aligned}$

With $n=99$ then the value of the sum is $\boxed{100!-1}$ .