9. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
a_{0}=2, a_{1}=3, a_{2}=6 \text {, }
$$

and for $n \geqslant 3$, we have
$$
a_{n}=(n+4) a_{n-1}-4 n a_{n-2}+(4 n-8) a_{n-3} \text {. }
$$

Find the general term of $\left\{a_{n}\right\}$.
Certainly, here is the translation while preserving the original formatting:

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9. Clearly,
$$
\begin{array}{l}
a_{n}-n a_{n-1} \\
=4\left(a_{n-1}-(n-1) a_{n-2}\right)-4\left(a_{n-2}-(n-2) a_{n-3}\right) .
\end{array}
$$

Let \( b_{n}=a_{n}-n a_{n-1} \). Then
$$
b_{n+1}=4 b_{n}-4 b_{n-1}, \quad b_{1}=1, \quad b_{2}=0 \text {. }
$$

Thus, \( b_{n}=2^{n}\left(1-\frac{n}{2}\right) \)
$$
\begin{aligned}
\Rightarrow & a_{n}-2^{n}=n a_{n-1}-n \cdot 2^{n-1} \\
\quad= & n\left(a_{n-1}-2^{n-1}\right) .
\end{aligned}
$$

Therefore, \( a_{n}=n!+2^{n} \).