An algebraic explanation. The fermionic path integral is based on a representation of the operators $a$ and $a^{\dagger}$ in terms of a <PERSON> variable $\alpha$ according to $a\rightarrow\alpha$ and $a^{\dagger}\rightarrow\partial_{\alpha}.$ The wave function $\psi\left(\alpha\right)=c_{0}\alpha+c_{1}$ is a function of $\alpha$.
The most general linear transformation from the particle number representation to the $\alpha$ representation is $\psi\left(\alpha\right)=\sum_{n=0}^{1}U_{\alpha,n}c_{n},$ i.e., it formally can be written as a matrix $U$ with a <PERSON> variable as index and a discrete index. This seems complicated, but $U_{\alpha,n}=\alpha x_{n}+y_{n}$ can be expanded in $\alpha$ and is determined by the four real or complex numbers $x_{0},x_{1},y_{0},y_{1}.$ To find out how the trace of a matrix $W_{m,n}$ looks like in the $\alpha$ representation the inverse $V_{n,\alpha}=\alpha p_{n}+q_{n}$ of the matrix $U$ is needed. 364
We thus require $$\sum_{n}U_{\alpha,n}V_{n,\beta}=\sum_{n}\left(\alpha\beta x_{n}p_{n}+\alpha x_{n}q_{n}+\beta y_{n}p_{n}+y_{n}q_{n}\right)=\beta-\alpha=\delta\left(\alpha-\beta\right).$$
The $O\left(\alpha\beta\right)$, $O\left(\alpha\right)$, $O\left(\beta\right)$ and $O\left(1\right)$ terms determine the four coefficients $p_{0},p_{1},q_{0},q_{1}$ of $V$. The result is $p_{n}=\left(\lambda x_{1},-\lambda x_{0}\right)$ and $q_{n}=\left(\lambda y_{1},-\lambda y_{0}\right)$ with $\lambda=1/\left(x_{1}y_{0}-x_{0}y_{1}\right).$
To get the trace one also needs the completeness relation in the particle number representation, which takes the form
$$\int\mathrm{d}\alpha V_{m,\alpha}U_{-\alpha,n}=\int\mathrm{d}\alpha\left(\alpha p_{m}+q_{m}\right)\left(-\alpha x_{n}+y_{n}\right)=p_{m}y_{n}-q_{m}x_{n}=\delta_{m,n}.$$
This equation essentially also is purely algebraic and easy to verify with $\int\mathrm{d}\alpha=\partial_{\alpha}$ and the the explicit solution for $p$ and $q$. Simply check the cases $mn\in\left{ 00,01,10,11\right} $. The minus sign is essential.
To get the expression for the trace insert this $\delta_{m,n}$ into $$\sum_{m,n}\delta_{m,n}W_{n,m}=\int\mathrm{d}\alpha\sum_{mn}U_{-\alpha,n}W_{nm}V_{m,\alpha}=\int\mathrm{d}\alpha W_{-\alpha,\alpha}.$$ 262