7. Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and it satisfies
$$
S_{n+1}-5 S_{n}-4 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

Then the last four digits of $a_{2 \text { on }}$ are $\qquad$
7.8124.

Given $S_{n+1}-5 S_{n}-4 n-4=0$, and
$$
S_{n}-5 S_{n+1}-4 n=0 \text {, }
$$

subtracting the two equations yields
$$
\begin{array}{l}
a_{n+1}-5 a_{n}-4=0 \\
\Rightarrow a_{n+1}+1=5\left(a_{n}+1\right) \\
\Rightarrow a_{n}=5^{n}-1 .
\end{array}
$$

Notice that, when $k \in \mathbf{N}_{+}$,
$$
\begin{aligned}
5^{4 k+1} & \equiv 3125\left(\bmod 10^{4}\right), \\
5^{4 k+2} & \equiv 5625\left(\bmod 10^{4}\right), \\
5^{4 k+3} & \equiv 8125\left(\bmod 10^{4}\right), \\
5^{4 k+4} & \equiv 625\left(\bmod 10^{4}\right) .
\end{aligned}
$$

Thus, $a_{2011} \equiv 5^{4 \times 502+3}-1 \equiv 8124\left(\bmod 10^{4}\right)$.