8. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{2010}=4$, the function
$$
f(x)=x\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{2010}\right) .
$$

Then the equation of the tangent line to the function $y=f(x)$ at the point $(0,0)$ is _. $\qquad$
8. $y=2^{2010} x$.
$$
\text { Let } g(x)=(x-a_{1})(x-a_{2}) \cdots(x-a_{2010}) \text {. }
$$

Then $f(x)=x g(x)$.
Since $f^{\prime}(x)=g(x)+x g^{\prime}(x)$, we have,
$$
\begin{array}{l}
f^{\prime}(0)=g(0)=a_{1} a_{2} \cdots a_{2010} \\
=\left(a_{1} a_{2010}\right)^{\frac{2010}{2}}=2^{2010} .
\end{array}
$$

Therefore, the equation of the tangent line at the point $(0,0)$ is
$$
y=2^{2010} x \text {. }
$$