Answer
See below.
Work Step by Step
1. For $n=1$, we have $LHS=1(1!)=1$ and $RHS=(1+1)!-1=1=LHS$, thus P(1) is true.
2. Assume P(k) ($k\gt1$) is true, we have:
$\sum_{i=1}^{k}i(i!)=(k+1)!-1$
3. For $n=k+1$, we have:
$LHS=\sum_{i=1}^{k}i(i!)+(k+1)(k+1)!\\
=(k+1)!-1+(k+1)(k+1)!\\
=(k+1)!(k+2)-1=(k+2)!-1=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.