Derivation of Grassmann valued functional

I'm trying to evaluate $$\frac{\delta}{\delta \eta(x)}e^{-\int dz \theta^*(z)\eta(z)}$$ Where $\theta^*(x)$ and $\eta(x)$ are Grassmann valued functions. The context of the functional is in term of functions that can be expressed in a basis of orthonormal functions $\{f_n(x)\}$ such that $$\eta(x)=\sum_n\eta_n f_n(x)$$ where $\eta_n$ is Grassmann number, which means that for any two Grassmann numbers, $a$, $b$ then $ab=-ba$ and so $a^2=0$, etc.

The definition of the derivative is such that $$\frac{\delta}{\delta \eta(x)}\eta(y)=\delta(x-y)$$ With that it is easy to show that $$\frac{\delta}{\delta \eta(x)}e^{-\int dz \theta^*(z)\eta(z)}=\theta^*(x)$$ However, this should be also equal to $$\theta^*(x)e^{-\int dz \theta^*(z)\eta(z)}$$ Which should be possible to show by noticing the property that $a^2=0$ (I think), so by a Taylor expansion $$\theta^*(x)=^?\theta^*(x)-\int \theta^*(x) \theta^*(z)\eta(z)dz=\theta^*(x)e^{-\int dz \theta^*(z)\eta(z)}$$ But I'm failing to see how the integral is zero. I tried working out whether $\theta^*(x)\theta^*(z)$ is zero alone by expanding the functions in the basis: $$\theta^*(x)\theta^*(z)=\sum_n \theta^2_nf^*_n(x)f^*_n(y)+\sum_{n<m}\theta_n \theta_m (f^*_n(x)f^*_m(y)-f^*_m(x)f^*_n(y))$$ The only properties of the $f$ I know are the orthogonality: $$\int dx f^*_m(x)f_n(x)=\delta_{mn}$$ and the completeness condition $$\sum_n f^*_n(x)f_n(y)=\delta(x-y)$$ But I don't think I can use those.

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maybe better the physics version of stack exchange. Your first formula with just \theta^{\ast}(x) is wrong. The second with also the exponential is correct. –  Abdelmalek Abdesselam Feb 20 at 21:31