Question 132, Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=1, a_{n+1}=\frac{a_{n}}{(n+1)\left(a_{n}+1\right)}\left(n \in Z^{+}\right)$, try to find the value of $\lim _{n \rightarrow+\infty} n!\cdot a_{n}$.
The 132nd question, answer: According to the conditions, we know:
$$
\begin{array}{l}
\frac{n+1}{a_{n+1}}=\frac{1}{a_{n}}+1 \\
\Rightarrow \frac{1}{(n+1) a_{n+1}}=\frac{1}{a_{n}}+1 \\
\Rightarrow \frac{1}{(n+1)!a_{n+1}}=\frac{1}{n!a_{n}}+\frac{1}{n!}
\end{array}
$$

Thus, we know:
$$
\begin{array}{l}
\frac{1}{n!a_{n}}=\frac{1}{1!\cdot a_{1}}+\sum_{i=1}^{n-1}\left(\frac{1}{(1+1)!a_{1+1}}-\frac{1}{11 a_{1}}\right)=1+\sum_{i=1}^{n-1}\left(\frac{1}{11}\right) \\
\Rightarrow n!a_{n}=\frac{1}{1+\sum_{i=1}^{n-1}\left(\frac{1}{n}\right)} \\
\Rightarrow \lim _{n \rightarrow+\infty} n!\cdot a_{n}=\lim _{n \rightarrow+\infty} \frac{1}{1+\sum_{i=1}^{n-1}\left(\frac{1}{n}\right)}=\frac{1}{e}
\end{array}
$$

Thus, we know $\lim _{n \rightarrow+\infty} n!\cdot a_{n}=\frac{1}{e}$.