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Let's have grassmannian numbers $\theta_{i}$: $$ \theta_{i}\theta_{j} = -\theta_{j}\theta_{i}, \quad \theta_{i} x = x\theta_{i}, $$ where x is just ordinary number.

How to understand (or how to show if it isn't the definition) that for two sets of grassmannian numbers $\theta_{i}, \eta_{j}$ $$ \int e^{\eta_{i}A_{ij}\eta_{j}} d\eta = \sqrt{det A_{ij}} ? \qquad (1) $$ I know that for grassmannian numbers the integration is equal to the differentiation, but I don't understand how to apply this statement for my question.

Addition.

Clearly it may be proved in the following way. First we "diagonalize" $A_{ij}$ to the form $$ A_{ij} = \begin{pmatrix} 0 & \lambda_{1} & 0 & 0 & ... & 0 & 0 \\ -\lambda_{1} & 0 & 0 & 0 & ... & 0 & 0 \\ 0 & 0 & 0 & \lambda_{2} & \dots & 0 & 0 \\ 0 & 0 & - \lambda_{2} & 0 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\ 0 & 0& 0& 0 & \dots & 0 & \lambda_{n} \\ 0 & 0& 0& 0 & \dots & -\lambda_{n} & 0\end {pmatrix} . $$ Then I can expand the exponentials as $$ e^{\eta_{i}A_{i, j}\eta_{j}} = 1 + \eta_{i}A_{i, i + 1}\eta_{j} + ... + \eta_{1}A_{12}\eta_{2}...\eta_{n - 1}A_{n - 1, n}\eta_{n}. $$ Finally, I use properties $\int 1 d \eta = 0, \quad \int \eta d \eta = 1$, so I will get $$ \int e^{\eta_{i}A_{ij}\eta_{j}}d\eta = \prod_{i = 1}^{\frac{n}{2}}\lambda_{i} = \sqrt{det A_{ij}}. $$ Can you check it?

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If you are integrating over an even dimensional Grassmann algebra than your reasoning is correct (to the best of my understanding you can only bring even dimensional skew-symmetric matrices to the block diagonal form you give). In this case your exponent can always be written in the form you give (lhs eq.1) where $A_{ij}$ is an $N\times N$ skew-symmetric matrix ($N=2m$ is even). This is true because one could write a general matrix as $M_{ij}=(M_{ij})_{sym}+(M_{ij})_{anti-sym}$ and the symmetric part would always vanish due to the Grassmann nature of the numbers ($\theta_i\theta_i=0$ and $\theta_i\theta_j=-\theta_j\theta_i$). If you want a more in dept derivation of the result you are asking about you can check my answer on a similar question here.

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There's a problem with your last step. Even if $n$ is even so that the product makes sense, it's still unlikely that you'll get $\sqrt{\det A}$.

You might want to take a look at my answer to a similar question here.

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