Example 4.5 For a fixed natural number $k$, find the ordinary generating function of the sequence $\left\{S_{2}(n, k)\right\}_{n \geqslant 0}$.
Let

Then
$$
\begin{array}{c}
g_{k}(t)=\sum_{n=0}^{\infty} S_{2}(n, k) t^{n}, \\
g_{k}(t)=\sum_{n=k}^{\infty} S_{2}(n, k) t^{n} \\
=\sum_{n=k}^{\infty}\left[S_{2}(n-1, k-1)+k S_{2}(n-1, k)\right] t^{n} \\
=t \sum_{n=k}^{\infty} S_{2}(n-1, k-1) t^{n-1}+ \\
k t \sum_{n=k+1}^{\infty} S_{2}(n-1, k) t^{n-1} \\
=t \sum_{n=k-1}^{\infty} S_{2}(n, k-1) t^{n}+k t \sum_{n=k}^{\infty} S_{2}(n, k) t^{n} \\
=t \sum_{n=0}^{\infty} S_{2}(n, k-1) t^{n}+k t \sum_{n=0}^{\infty} S_{2}(n, k) t^{n} \\
=t g_{k-1}(t)+k t g_{k}(t),
\end{array}
$$

So
$$
(1-k t) g_{k}(t)=t g_{k-1}(t).
$$
$$
\begin{aligned}
g_{k}(t) & =\frac{t}{1-k t} g_{k-1}(t) \\
& =\frac{t}{1-k t} \cdot \frac{t}{1-(k-1) t} g_{k-2}(t) \\
& =\cdots \\
& =\frac{t}{1-k t} \cdot \frac{t}{1-(k-1) t} \cdots \cdots \frac{t}{1-2 t} \cdot g_{1}(t) .
\end{aligned}
$$

Since
$$
g_{1}(t)=\sum_{n=1}^{\infty} S_{2}(1,1) t^{n}=\sum_{n=1}^{\infty} t^{n}=\frac{t}{1-t},
$$

Therefore
$$
\begin{aligned}
g_{k}(t) & =\frac{t}{1-k t} \cdot \frac{t}{1-(k-1) t} \cdot \cdots \cdot \frac{t}{1-2 t} \cdot \frac{t}{1-t} \\
& =\frac{t^{k}}{(1-t)(1-2 t) \cdots(1-k t)} .
\end{aligned}
$$