8. The unit digit of $\left[\frac{10^{10000}}{10^{100}+9}\right]$ is
8. 1 .

Notice that,
$$
\frac{10^{10000}}{10^{100}+9}=\frac{\left(10^{100}\right)^{100}-3^{200}}{10^{100}+9}+\frac{3^{200}}{10^{100}+9} \text {. }
$$

And $\left(10^{100}\right)^{100}-3^{200}=\left[\left(10^{100}\right)^{2}\right]^{50}-\left(9^{2}\right)^{50}$, so $\left[\left(10^{100}\right)^{2}-9^{2}\right] \mid\left(10^{10000}-9^{200}\right)$.
Since $\left(10^{100}+9\right) \mid \left[\left(10^{100}\right)^{2}-9^{2}\right]$, therefore, $\frac{10^{10000}-3^{200}}{10^{100}+9}$ is an integer.
By $\frac{3^{200}}{10^{100}+9}=\frac{9^{100}}{10^{100}+9}<1$, we know
$$
\begin{array}{l}
{\left[\frac{10^{10000}}{10^{100}+9}\right]=\frac{10^{10000}-3^{200}}{10^{100}+9}} \\
=\frac{10^{10000}-9^{100}}{10^{100}+9}=\frac{10^{10000}-81^{50}}{10^{100}+9},
\end{array}
$$

where the unit digit of the denominator is 9, and the unit digit of the numerator is 9, so the unit digit of the quotient is 1.