Note that \[\tan (90^\circ - x) \tan x = \frac{\sin (90^\circ - x)}{\cos (90^\circ - x)} \cdot \frac{\sin x}{\cos x} = \frac{\cos x}{\sin x} \cdot \frac{\sin x}{\cos x} = 1,\]so \begin{align*} &\tan 10^\circ \tan 20^\circ \tan 30^\circ \tan 40^\circ \tan 50^\circ \tan 60^\circ \tan 70^\circ \tan 80^\circ \\ &= (\tan 10^\circ \tan 80^\circ) (\tan 20^\circ \tan 70^\circ) (\tan 30^\circ \tan 60^\circ) (\tan 40^\circ \tan 50^\circ) \\ &= \boxed{1}.\end{align*}