7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=1, a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}=0 \text {. }
$$

Then $a_{2008}=$ $\qquad$
7.2007!.
Dividing both sides of the recurrence relation by $a_{n+1} a_{n}$, we get
$$
\frac{a_{n+2}}{a_{n+1}}-\frac{a_{n+1}}{a_{n}}=1 \text {. }
$$

Thus, the sequence $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1. Therefore, $\frac{a_{n+1}}{a_{n}}=n$.
Hence, $a_{2008}=2007!$.