For example, $73 \leqslant n \in \mathbf{N}$, let $S$ be the set of all non-empty subsets of the set $\{2,3, \cdots, n\}$. For each $S_{i} \subseteq S, i=1,2, \cdots, 2^{n-1}-1$, let $p_{i}$ be the product of all elements in $S_{i}$. Find $p_{1}+p_{2}+\cdots+p_{2^{n-1}-1}$.
When $n=3$, it is easy to get
$$
\begin{array}{l}
p_{1}+p_{2}+p_{3} \\
=2+3+2 \times 3=11=\frac{4!}{2}-1,
\end{array}
$$

When $n=4$, we have
$$
\begin{array}{l}
p_{1}+p_{2}+p_{5}+p_{4}+p_{5}+p_{6}+p_{7} \\
=11+4+4(2+3+2 \cdot 3)=59 \\
=\frac{5!}{2}-1 ;
\end{array}
$$

From this, we conjecture
$$
\begin{array}{l}
p_{1}+p_{2}+\cdots+p_{n}^{n-1}-1 \\
=\frac{(n+1)!}{2}-1(n \in \mathbf{N} \text { and } n \geqslant 3) .
\end{array}
$$

We will prove that equation (*) holds using mathematical induction. When $n=3$, the conclusion is obviously correct;
Assume that for $n=k(k \geqslant 3)$, equation (*) is correct, i.e.,
$$
p_{1}+p_{2}+\cdots+p_{k}^{k-1}-1=\frac{(k+1)!}{2}-1 \text {. }
$$

By adding an element $k+1$ to the set $\{2,3, \cdots, k\}$, the product of its elements increases by
$$
\begin{array}{l}
(k+1) \cdot \frac{(k+1)!}{2} . \\
\therefore p_{1}+p_{2}+\cdots+p_{k+1}^{k}-1 \\
=\frac{(k+1)!}{2}-1+(k+1) \cdot \frac{(k+1)!}{2} \\
=\frac{(k+2)!}{2}-1 .
\end{array}
$$

Therefore, for all $3 \leqslant n \in \mathbf{N}$, we have
$$
p_{1}+p_{2}+\cdots+p_{n}^{n-1}-1=\frac{(n+1)!}{2}-1 .
$$