Without limitation of generality, suppose that the left lower cell $A$ and the right upper cell $Z$ were cut out. First, let us see in how many ways rooks can be placed on a $n \times n$ chessboard. There are $n!$ ways, since for a rook, there are $n$ options in the first column, $n-1$ option in the second column, and so on. Consider the arrangements of rooks, in which one of them is located in cell $A$. There are $(n-1)!$ such arrangements, and analogously, $(n-1)!$ ways, for when a rook occupies cell $Z$. In the expression $n!-2(n-1)!$ positions where rooks occupy cells $A$ and $Z$ are rejected twice. There are $(n-2)!$ such arrangements. Thus, the correct expression would be:
$$n!-2(n-1)!+(n-2)!=(n-2)!(n(n-1)-2(n-1)+1)=(n-2)!(n^2-3n+3).$$