Answer
See below.
Work Step by Step
Let the four consecutive numbers be $n, n+1, n+2, n+3$, we have:
case1: it is even, $n=2k$, thus $n(n+2)=2k(2k+2)=4k(k+1)$, as one of $k$ or $k+1$ must be even, $n(n+2)$ is divisible by 8;
case2: it is odd, $n=2k+1$, thus
$(n+1)(n+3)=(2k+2)(2k+4)=4(k+1)(k+2)$, as one of $k+1$ or $k+2$ must be even, $(n+1)(n+3)$ is divisible by 8;
thus, in any case, the product of any four consecutive numbers is divisible by 8.
