15. (13 points) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}$ $-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.


15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}$
$$
\begin{array}{l}
-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1) \cdot \\
\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots+4 \cdot 3 \cdot\left(a_{2}-\right. \\
\left.a_{1}\right)=(n+2)!, \\
\text { Therefore, } a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right) \\
\quad=1+2!+3!+\cdots+n!(n \geqslant 1) .
\end{array}
$$

Therefore, $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right)$

Given $a_{1}=1, a_{2}=3, a_{3}=9, a_{4}=33, a_{5}=153$, at this point 153 is divisible by 9.

When $m \geqslant 5$, $a_{n}=a_{5}+\sum_{k=6}^{m} k!$, and when $k \geqslant 6$, $k!$ is divisible by 9.

Thus, when $m \geqslant 5$, $a_{n}$ is divisible by 9. Therefore, the smallest value of $n$ is 5.