Example 1 In the sequence $\left\{a_{n}\right\}$, $a_{0}=1, a_{1}=2$, and $n(n+1) a_{n+1}=n(n-1) a_{n}-(n-2) a_{n-1}, n \geqslant 1$. Find $a_{n}(n \geqslant 2)$.
Solve: From $1 \cdot 2 a_{2}=1 \cdot 0 \cdot a_{1}-(-1) a_{0}$, we get $a_{2}=\frac{1}{2!}$  
From $2 \cdot 3 a_{3}=2 \cdot 1 a_{2}-0 \cdot a_{1}$, we get $a_{3}=\frac{1}{3!}$  
From $3 \cdot 4 a_{4}=3 \cdot 2 a_{3}-1 \cdot a_{2}$, we get $a_{4}=\frac{1}{4!}$  

Thus, we propose the conjecture $a_{n}=\frac{1}{n!}, n \geqslant 2$  
This is easy to prove by mathematical induction. In fact, it has been verified for $n=2,3,4$. Assume  
$$
a_{k}=\frac{1}{k!}, k=2,3, \cdots, n
$$

Then $k(k+1) a_{k+1}=k(k-1) a_{k}-(k-2) a_{k-1}, k \geqslant 3$  
$$
=\frac{k(k-1)}{k!}-\frac{k-2}{(k-1)!}=\frac{1}{(k-1)!}
$$

So $a_{k+1}=\frac{1}{(k+1)!}, k \geqslant 3$, which means the conjecture holds for $n=k+1$, thus $a_{n}=\frac{1}{n!}, n \geqslant 2$