jmc

We can re-write the given expression as $(16\times 17\times 18{)}^{17}\times 17\times 18^2$. First, we find the units digit of $(16\times 17\times 18{)}^{17}$. The units digit of $16\times 17\times 18$ is that of $6\times 7\times 8,$ or that of $2\times 8$, or $6$. When raised to any perfect power, a positive integer ending in $6$ will still end in $6$, so $(16\times 17\times 18{)}^{17}$ has a units digit of $6$. Now, we need to find the units digit of $6\times 17\times 18^2$, or the units digit of $2\times 18^2$, which is $\boxed{8}$.