I am trying to prove that if A is a $m x m$ real matrix and $B=\begin{bmatrix}0 & A\\-A^t & 0\end{bmatrix}$, then Pfaffian(B)= $(-1)^{m(m-1)/2}$ det(A).
I know that Pfaffian$(B)^2=$det(B), so in this case Pfaffian$(B)^2=det(A)^2$, so Pfaffian(B)=$\pm$det(A) but I am not sure how to proceed next. Thank you!