jmc

All the tens digits of the product will be irrelevant to finding the units digit. Thus, we are searching for the units digit of $(2\cdot 4\cdot 6\cdot 8)\cdot (2\cdot 4\cdot 6\cdot 8)\cdot (2\cdot 4\cdot 6\cdot 8)\cdot ...$ There will be $10$ groups of $4$ numbers. The number now can be rewritten as $(2\cdot 4\cdot 6\cdot 8{)}^{10}$. Multiplying, we get $384^{10}$. Again, we can disregard the tens and hundreds digit of $384$, since we only want the units digit of the number, leaving $4^{10}$. Now, we try to find a pattern to the units digit of $4^n$. To compute this quickly, we once again discard all tens digits and higher. $4^1=4$. $4^2=4\cdot 4=1\underline{6}$, discard the $1$. $4^3=1\cdot 4=\underline{4}$ $4^4=4\cdot 4=1\underline{6}$, discard the $1$. $4^5=1\cdot 4=\underline{4}$ Those equalities are, in reality, congruences $\mathrm{mod}10$. Thus, the pattern of the units digits is $\{4,6,4,6,4,6,4,6,4,6\}$. The cycle repeats so that term $n$ is the same as term $n+2$. The tenth number in the cycle is $6$, giving an answer of $\boxed{6}$.