There are $ 8\cdot 8 = 64 $ ordered pairs that can represent the top numbers on the two dice. Let $ m $ and $ n $ represent the top numbers on the dice. Then $ mn > m+n $ implies that $ mn - m - n > 0 $, that is, $$1 < mn - m - n + 1 = (m-1)(n-1).$$This inequality is satisfied except when $ m=1 $, $ n=1 $, or when $ m=n=2 $. There are 16 ordered pairs $ (m,n) $ excluded by these conditions, so the probability that the product is greater than the sum is \[ \frac{64-16}{64} = \frac{48}{64} = \boxed{\frac{3}{4}}.\]