$3 \cdot 2$ Find $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+(n-1)(n-1)!+n \cdot n$ !, where $n!=n(n-1)(n-2) \cdots 2 \cdot 1$.
[Solution] Since $k \cdot k!=(k+1)!-k!\quad k=1,2,3, \cdots$
Therefore,
$$\begin{aligned}
& 1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+(n-1)(n-1)!+n \cdot n! \\
& =(2!-1!)+(3!-2!)+\cdots+[n!-(n-1)!]+[(n+1)!- \\
& n!] \\
= & (n+1)!-1 .
\end{aligned}$$