14. Summation of Factorials. What is the sum

$$
1(1!)+2(2!)+3(3!)+\cdots+n(n!) ?
$$
14. We have

$$
\begin{aligned}
& 1(1!)+2(2!)+3(3!)+\ldots+(n-1)[(n-1)!]+n(n!)= \\
& \quad=2(1!)+3(2!)+4(3!)+\ldots+n[(n-1)!]+(n+1)(n!)- \\
& \quad-1!-2!-3!-\ldots-(n-1)!-n!=(n+1)!-1
\end{aligned}
$$