$$
\begin{aligned}
a_{n} & =(n+2) a_{n-1}=(n+2)(n+1) a_{n-2} \\
& =\cdots=(n+2)(n+1) \cdots 3 \cdot a_{0} \\
& =(n+2)(n+1) \cdots 3 \cdot 2=(n+2)!
\end{aligned}
$$

Solving using the iterative method, continuously applying the recursive relation, we have
Example 2 In the sequence $\left\{a_{n}\right\}$, $a_{0}=2$, and $a_{n}=(n+2) a_{n-1}, n \geqslant 1$, find $a_{n}$.