I am working in tilings using Pfaffian. There is a basic property, namely:
Let $ B$ be a $n\times n$ matrix. Let $$ A = \begin{pmatrix} 0 & B\\ -B^T & 0 \end{pmatrix}$$
then
$$\operatorname{Pf}(A)=(-1)^{\frac{n(n-1)}{2} }\det(B)$$
thanks for the hints.
The property $\operatorname{Pf}(A)^2=\det(A)$ gives me $\operatorname{Pf}(A)=\pm\sqrt{\det(A)}$. With the $n$ column exchanges we get
$$\det (A) = (-1)^n\det \begin{pmatrix}
B & 0\\
0& -B^T
\end{pmatrix}=(-1)^n\det(B)\cdot (-1)^n\det (B^T)$$
But somewhere along I just miss something important :(