jmc

If $k$ is a positive even integer, then $$9^k=81^{k/2}=\stackrel{k/2\text{ times}}{\overbrace{(81)(81)\cdots (81)}},$$so $9^k$ has a units digit of 1. Since 8 is even, $8^7$ is even. Therefore, $9^{8^7}$ has a units digit of $\boxed{1}$.