10. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{3}{2}$, and $a_{n}=\frac{3 n a_{n-1}}{2 a_{n-1}+n-1}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) For all positive integers $n$, the inequality $a_{1} a_{2} a_{3} \cdots a_{n}<\lambda \cdot n!$ always holds, try to find the minimum value of the positive integer $\lambda$.
10. Solution: (1) From $a_{n}=\frac{3 n a_{n-1}}{2 a_{n-1}+n-1} \Rightarrow \frac{a_{n}}{n}=\frac{3 a_{n-1}}{2 a_{n-1}+n-1} \Rightarrow \frac{n}{a_{n}}=\frac{2}{3}+\frac{n-1}{3 a_{n-1}} \Rightarrow \frac{n}{a_{n}}-1=\frac{1}{3}\left(\frac{n-1}{a_{n-1}-1}-1\right)$,
$\therefore$ the sequence $\left\{\frac{n}{a_{n}}-1\right\}$ is a geometric sequence with a common ratio of $\frac{1}{3}$, and the first term is $\frac{1}{a_{1}}-1=-\frac{1}{3}, \frac{n}{a_{n}}-1=-\left(\frac{1}{3}\right)^{n} \Rightarrow a_{n}=\frac{n}{1-\frac{1}{3^{n}}}$.
(2) $a_{1} a_{2} a_{3} \cdots a_{n}1 \Leftrightarrow\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^{2}}\right) \cdots$ $\left(1-\frac{1}{3^{n}}\right)>\frac{1}{\lambda}$ always holds.
First, we prove the conclusion $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^{2}}\right) \cdots\left(1-\frac{1}{3^{n}}\right)>1-\sum_{k=1}^{n} \frac{1}{3^{k}}$ (using mathematical induction) (omitted). And $1-\sum_{k=1}^{n} \frac{1}{3^{k}}=\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}\right)^{n}, \lim _{n \rightarrow \infty}\left(1-\sum_{k=1}^{n} \frac{1}{3^{k}}\right)=\frac{1}{2}, \therefore \frac{1}{2} \geqslant \frac{1}{\lambda} \Rightarrow \lambda \geqslant 2$, i.e., $\lambda_{\text {min }}=2$.