Prove that the Pfaffian satisfies $\text{Pf}(MAM^T)=\det(M)\text{Pf}(A)$ 
Show that $$\text{Pf} MAM^T = \text{det}M \cdot \text{Pf} A$$ for any matrix $M$ and antisymmetric $A$.

Attempt: $$\text{Pf} MAM^T = \frac{1}{2^N N!} \epsilon_{\alpha_1 \dots \alpha_{2N}} (MAM^T)_{\alpha_1 \alpha_2} \dots (MAM^T)_{\alpha_{2N-1} \alpha_{2N}} = \frac{1}{2^N N!}\epsilon_{\alpha_1 \dots \alpha_{2N}} M_{\alpha_1 \sigma_1} A_{\sigma_1 \delta_1} (M^T)_{\delta_1 \alpha_2} \dots M_{\alpha_{2N-1} \sigma_{2N-1}} A_{\sigma_{2N-1} \delta_{2N-1}} (M^T)_{\delta_{2N-1} \alpha_{2N}}  $$ while $$\text{det} M = \epsilon_{\beta_1 \dots \beta_{2N}} M_{1, \beta_1} \dots M_{2N, \beta_{2N}}$$ and $$\text{Pf}A = \frac{1}{2^N N!} \epsilon_{\gamma_1 \dots \gamma_{2N}} (A)_{\gamma_1 \gamma_2} \dots (A)_{\gamma_{2N-1} \gamma_{2N}}$$ 
Working with the terms on the r.hs I see that $$\text{Pf}A \cdot \det M = \frac{1}{2^N N!} \epsilon_{\beta_1 \dots \beta_{2N}} \epsilon_{\gamma_1 \dots \gamma_{2N}} M_{1, \beta_1} \dots M_{2N, \beta_{2N}}(A)_{\gamma_1 \gamma_2} \dots (A)_{\gamma_{2N-1} \gamma_{2N}}$$ I don't see a way to proceed - is there perhaps another definition of $det$ I should use or can I argue based on these diagrammatic forms below?

 A: Here is an approach using (possibly complex) Grassmann variables and Berezin integration$^1$. 


*

*Define the Pfaffian of a (possibly complex) antisymmetric matrix $A^{ij}=-A^{ji}$ (in $n$ dimensions$^2$) as
$$ \begin{align}{\rm Pf}(A)~:=~&\int \!d\theta_n \ldots d\theta_1~ e^{\frac{1}{2}\theta_i A^{ij}\theta_j}\cr
~\stackrel{(5)}{=}~&\frac{\partial}{\partial \theta_n} \ldots \frac{\partial}{\partial \theta_1} e^{\frac{1}{2}\theta_i A^{ij}\theta_j}\cr
~=~&\frac{1}{n!}\epsilon_{i_1\ldots i_n} \frac{\partial}{\partial \theta_{i_n}} \ldots \frac{\partial}{\partial \theta_{i_1}} e^{\frac{1}{2}\theta_i A^{ij}\theta_j}.\end{align} \tag{1}$$

*If we make a change of coordinates 
$$ \theta^{\prime}_j~=~\theta_i M^i{}_j,\tag{2} $$
the chain rule becomes
$$ \frac{\partial}{\partial \theta_i}~=~M^i{}_j\frac{\partial}{\partial \theta^{\prime}_j} .\tag{3} $$

*Therefore OP's first equation follows from
$$\begin{align}
{\rm Pf}(MAM^T)
&~~\stackrel{(1)}{=}~\frac{1}{n!}\epsilon_{i_1\ldots i_n} \frac{\partial}{\partial \theta_{i_n}} \ldots \frac{\partial}{\partial \theta_{i_1}} e^{\frac{1}{2}\theta_i  M^i{}_k A^{k\ell}M^j{}_{\ell}\theta_j}\cr
&\stackrel{(2)+(3)}{=}~\frac{1}{n!}\epsilon_{i_1\ldots i_n}M^{i_1}{}_{j_1}\ldots M^{i_n}{}_{j_n} \frac{\partial}{\partial \theta^{\prime}_{j_n}} \ldots \frac{\partial}{\partial \theta^{\prime}_{j_1}}  e^{\frac{1}{2}\theta^{\prime}_i A^{ij}\theta^{\prime}_j}   \cr
&~~=~\epsilon_{i_1\ldots i_n}M^{i_1}{}_{1}\ldots M^{i_n}{}_{n} ~\frac{\partial}{\partial \theta^{\prime}_{n}} \ldots \frac{\partial}{\partial \theta^{\prime}_{1}}  e^{\frac{1}{2}\theta^{\prime}_i A^{ij}\theta^{\prime}_j} \cr
&~~\stackrel{(1)}{=}~{\rm Det}(M)~{\rm Pf}(A).\end{align}\tag{4} $$
$\Box$
--
$^1$ We use the sign convention that Berezin integration $$\int d\theta_i~\equiv~\frac{\partial}{\partial \theta_i}\tag{5} $$ is the same as differentiation wrt. $\theta_i$ acting from left. See e.g. this Phys.SE post.
$^2$ One may show that the Pfaffian vanishes in odd dimensions.
