Combinatorics

At the top row, you can mark 6 squares because you can't pick the two corners. At the bottom row, you can mark 5 squares because you can't mark the same column or the two corners.

In the second row, you have 6 choices, thus leaving us 5, then 4, then 3, and so on.

So our final is $6\times5\times6!=30\times720=\boxed{21600}$

Without the restriction, there are $8!$ ways. There are 4 corners, so there are $4\times 7!$ violations. Then we must add back in the cases where two opposite corner squares are occupied. There are $2\times 6!$ ways for this, so $8!-4\times 7! + 2\times 6! = 21600$ .