(14) (13 points) Let the set $I=\{1,2,3, \cdots, n\}\left(n \in \mathbf{N}^{*}\right)$, choose two non-empty subsets $A$ and $B$ of $I$, such that the smallest number in $B$ is greater than the largest number in $A$, denote the number of different selection methods as $a_{n}$, obviously $a_{1}=0, a_{2}=\mathrm{C}_{2}^{2}=1$.
(1) Find $a_{n}$;
(2) Let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}$, find $S_{n}$.
(14) (1) According to the problem, we have $a_{1}=0$.
When $n \geqslant 2$,
$$
\begin{aligned}
a_{n}= & \mathrm{C}_{n}^{2}+2 \mathrm{C}_{n}^{3}+3 \mathrm{C}_{n}^{4}+\cdots+(n-1) \mathrm{C}_{n}^{n} \\
= & \left(2 \mathrm{C}_{n}^{2}+3 \mathrm{C}_{n}^{3}+\cdots+n \mathrm{C}_{n}^{n}\right)-\left(\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}+\cdots+\mathrm{C}_{n}^{n}\right) \\
= & \left(\mathrm{C}_{n}^{1}+2 \mathrm{C}_{n}^{2}+3 \mathrm{C}_{n}^{3}+\cdots+n \mathrm{C}_{n}^{n}\right)-\left(\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}+\cdots+\mathrm{C}_{n}^{n}\right) \\
= & n\left(\mathrm{C}_{n-1}^{0}+\mathrm{C}_{n-1}^{1}+\mathrm{C}_{n-1}^{2}+\cdots+\mathrm{C}_{n-1}^{n-1}\right) \\
& -\left(\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{2}+\mathrm{C}_{n}^{3}+\cdots+\mathrm{C}_{n}^{n}\right) \\
= & n \times 2^{n-1}-\left(2^{n}-1\right) \\
= & n \cdot 2^{n-1}-2^{n}+1 .
\end{aligned}
$$

Since $a_{1}=0, a_{2}=1$ also satisfy the above formula, we have
$$
a_{n}=n \cdot 2^{n-1}-2^{n}+1\left(n \in \mathbf{N}^{*}\right) .
$$
(2) From (1), we get
$$
\begin{aligned}
S_{n}= & a_{1}+a_{2}+a_{3}+\cdots+a_{n} \\
= & \left(1 \times 2^{0}+2 \times 2^{1}+3 \times 2^{2}+\cdots+n \times 2^{n-1}\right) \\
& -\left(2^{1}+2^{2}+2^{3}+\cdots+2^{n}\right)+n,
\end{aligned}
$$

Let
$$
T_{n}=1 \times 2^{0}+2 \times 2^{1}+3 \times 2^{2}+\cdots+n \times 2^{n-1},
$$

then
$$
2 T_{n}=1 \times 2^{1}+2 \times 2^{2}+\cdots+n \times 2^{n},
$$
Subtracting (2) from (1) gives
$$
\begin{aligned}
-T_{n} & =1+2^{1}+2^{2}+\cdots+2^{n-1}-n \times 2^{n} \\
& =2^{n}-1-n \times 2^{n} \\
& =(1-n) \cdot 2^{n}-1,
\end{aligned}
$$

Thus, $T_{n}=(n-1) \cdot 2^{n}+1$. Therefore,
$$
\begin{aligned}
S_{n} & =(n-1) \cdot 2^{n}+1-\left(2^{n+1}-2\right)+n \\
& =(n-3) \cdot 2^{n}+n+3
\end{aligned}
$$