$3 \cdot 2$ Find the sum of $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+(n-1)(n-1)!+n \cdot n!$, 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,
$$
\begin{array}{l} 
1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\cdots+(n-1)(n-1)!+n \cdot n! \\
=(2!-1!)+(3!-2!)+\cdots+[n!-(n-1)!]+[(n+1)!- \\
\quad n!]
\end{array}
$$
$$
=(n+1)!-1 .
$$