= Solution
Choose the orientation of the <Berezin integral> so that
$$
D\eta\,D\bar\eta=d\eta_N\,d\bar\eta_N\cdots d\eta_1\,d\bar\eta_1,
\qquad\int D\eta\,D\bar\eta\prod_{i=1}^N(\bar\eta_i\eta_i)=1.
$$
Here the product has increasing $i$; this explicitly fixes the otherwise convention-dependent overall sign in the compact measure notation.
Since $A$ is a <diagonalizable matrix>, write $A=S\Lambda S^{-1}$ with $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_N)$. Make the independent changes $\eta=S\xi$ and $\bar\eta=\bar\xi S^{-1}$. The <Grassmann change-of-variables formula> gives the two factors $(\det S)^{-1}$ and $\det S$, so the complete measure is unchanged. The exponent becomes $\sum_i\lambda_i\bar\xi_i\xi_i$. Each summand is even and squares to zero, and the different even summands commute. Consequently,
$$
e^{\bar\eta A\eta}=\prod_{i=1}^N(1+\lambda_i\bar\xi_i\xi_i).
$$
Only the term containing every generator survives the <Berezin integral>. Hence the <Grassmann Gaussian integral> is
$$
\boxed{\int D\eta\,D\bar\eta\,e^{\bar\eta A\eta}
=\prod_{i=1}^N\lambda_i=\det A}.
$$
Zero eigenvalues give zero on both sides, so invertibility of $A$ is unnecessary. In fact the identity extends to all ordinary matrices: the top-degree coefficient of the exponential is the alternating <determinant> expansion. The assumption that $A$ is a <diagonalizable matrix> makes the proof especially transparent.
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