Put \(\Omega =\sum _{i<j}A_{ij}\xi _i\xi _j =\xi ^TA\xi/2\). In \(\Omega ^m\), a nonzero top word uses each generator exactly once. The even two-letter factors commute; every matching appears in \(m!\) orders. Reordering its word to \(\xi _1\cdots \xi _{2m}\) gives its matching sign. Therefore \[\frac {\Omega ^m}{m!} =\operatorname {Pf}(A)\xi _1\cdots \xi _{2m}.\] The top term of \(e^{-\Omega }\) is \((-1)^m\Omega ^m/m!\). Reversing \(2m\) generators has sign \((-1)^{m(2m-1)}=(-1)^m\); the two signs cancel. This proves the integral identity, without an inverse.

For the nonvanishing assertion, regard \(A\) as a nondegenerate alternating form on the vector space dual to the generators. Choose a nonzero \(u\) and then \(v\) with \(A(u,v)=1\). The plane they span is nondegenerate, and its orthogonal complement for \(A\) is a direct complement on which \(A\) remains nondegenerate. Induction gives a basis of such pairs. In the dual basis \(\alpha _1,\beta _1,\ldots ,\alpha _m,\beta _m\), \[\Omega =\sum _{r=1}^m\alpha _r\beta _r,\qquad \frac {\Omega ^m}{m!}=\alpha _1\beta _1\cdots \alpha _m\beta _m\ne 0.\] An invertible change of generators multiplies a top word by a nonzero determinant. The displayed top coefficient is therefore nonzero in the original basis as well, proving \(\operatorname {Pf}(A)\ne 0\). The zero-dimensional case is \(1\).