II. (50 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a_{2}=1$, and $a_{n+1} a_{n-1}=a_{n}^{2}+n a_{n} a_{n-1}(n=2,3, \cdots)$.
(1) Find $a_{n}$;
(2) Find $\sum_{k=2}^{n} \frac{a_{k}}{(k-2)!}$.
(1) From $a_{n+1} a_{n-1}=a_{n}^{2}+n a_{n} a_{n-1}$, we can get
$$
\frac{a_{n+1}}{a_{n}}=\frac{a_{n}}{a_{n-1}}+n \text {. }
$$

Let $b_{n}=\frac{a_{n+1}}{a_{n}}$, thus, $b_{n}=b_{n-1}+n, b_{1}=\frac{a_{2}}{a_{1}}=1$.
$$
\begin{array}{l}
\text { Hence } b_{n}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)+b_{1} \\
=\sum_{i=1}^{n-1}(i+1)+1=\frac{n(n+1)}{2} .
\end{array}
$$

Then $a_{n}=\left(\prod_{i=1}^{n-1} \frac{a_{i+1}}{a_{i}}\right) a_{1}=\left(\prod_{i=1}^{n-1} b_{i}\right) a_{1}$
$$
=\prod_{i=1}^{n-1} \frac{i(i+1)}{2} \times 1=\frac{(n-1)!\cdot n!}{2^{n-1}} \text {. }
$$
$$
\begin{array}{l}
\text { (2) } \sum_{k=2}^{n} \frac{a_{k}}{(k-2)!}=\sum_{k=2}^{n} \frac{(k-1) k!}{2^{k-1}} \\
=\sum_{k=2}^{n} \frac{(k+1-2) k!}{2^{k-1}}=\sum_{k=2}^{n} \frac{(k+1)!}{2^{k-1}}-\sum_{k=2}^{n} \frac{k!}{2^{k-2}} \\
=\sum_{k=2}^{n} \frac{(k+1)!}{2^{k-1}}-\sum_{k=1}^{n-1} \frac{(k+1)!}{2^{k-1}}=\frac{(n+1)!}{2^{n-1}}-2 .
\end{array}
$$