Solution
= Solution
Use <mathematical induction> on $n\ge0$. For $n=0$, the sum is $0!\,0=0=1!-1$. Suppose it is $(n+1)!-1$ at $n$. Then
$$
\sum_{i=0}^{n+1}i!\,i=(n+1)!-1+(n+1)!(n+1)=(n+2)!-1.
$$
The induction is complete, proving \b[$\sum_{i=0}^n i!\,i=(n+1)!-1$]. The <factorial> identity $i!\,i=(i+1)!-i!$ also explains the cancellation directly.