Proof involving exponential of anticommuting operators Problem:
On page 23 of the book "Quarks, gluons and lattices" by Creutz, he defines a state
$$\langle\psi|=\langle 0|e^{bFc}e^{\lambda b^\dagger G c^\dagger}$$
where $\lambda$ is a number, $F, G$ are $N\times N$ symmetric matrices and $b, c$ are vectors whose components $b_m, c_m$ are operators such that their anticommutators satisfy
$$\{b_m^\dagger, b_n\}=\{c_m^\dagger, c_n\}=\delta_{mn}$$
with every other anti-commutator being zero, and the state $\langle 0 |$ such that
$$\langle 0 |c^\dagger_m=\langle 0 |b^\dagger_m=0$$
Creutz says that a straightfoward calculation can lead us to proving that
$$\langle\psi|b^\dagger=-\langle\psi|(F^{-1}-\lambda G)^{-1} c=-\langle\psi|(1-\lambda FG)^{-1}Fc$$

What I have so far:
For $N=1$, I have managed to prove this, but for $N>1$, I am struggling with how to show this. Something that I managed to prove is that
$$\langle\psi |b_k^\dagger=-\langle 0 | \sum_{n=1}^N F_{kn}c_n \prod_{m=1\neq k, l}^N \left(1+b_m F_{ml}c_l \right)e^{\lambda b^\dagger G c^\dagger}$$
which generalizes what I initially found for the $N=1$ case, which is $\langle\psi|b^\dagger=-\langle 0|Fc$. For this last result for the $N=1$ case, I then substituted $\langle 0 |=\langle \psi | e^{-\lambda b^\dagger G c^\dagger}e^{-bFc}$, and after some manipulations, I reached the desired result. Doing something similar for the $N>1$ case, I found it too difficult to get to a meaningful result (or even close to the final result).
Note that for the $N>1$ case, I have used that, for example,
$$\langle 0|e^{bFc}=\langle 0 |e^{\sum_{ij}b_i F_{ij} c_j}=\langle 0|\prod_{ij}(1+b_iF_{ij}c_j)$$
which is found on page 193 of Fradkin's book "Quantum Field theory: an integrated approach".
If anybody can give a hint or some guidance on this, or even provide with some reference that can help, it would be greatly appreciated.
 A: This answer will not correctly format on portrait mobile screens.
Let $f\equiv b_iF_{ij}c_j$ and $g=b_i^\dagger G_{ij}c_j^\dagger$; thus, $\langle\psi|\equiv\langle0|e^fe^{\lambda g}$. Using Einstein summation convention.
The following commutators will prove useful:
$$\begin{align}gb_m^\dagger&=b_i^\dagger G_{ij}c_j^\dagger b_m^\dagger&&\\&=-b_i^\dagger G_{ij}b_m^\dagger c_j^\dagger&\text{as $\left\{c_j^\dagger,b_m^\dagger\right\}=0$}&\\&=b_m^\dagger b_i^\dagger G_{ij}c_j^\dagger&\text{as $\left\{b_i^\dagger,b_m^\dagger\right\}=0$}&\\&=b_m^\dagger g\implies\left[g,b_m^\dagger\right]=0\end{align}$$
Similarly, $\left[g,b_m^\dagger\right]=\left[g,c_m^\dagger\right]=\left[f,b_m\right]=\left[f,c_m\right]=0$
$$\begin{align}fb_m^\dagger&=b_iF_{ij}c_jb_m^\dagger\\&=-b_iF_{ij}b_m^\dagger c_j&\text{as $\left\{c_j,b_m^\dagger\right\}=0$}&\\&=b_m^\dagger b_iF_{ij}c_j-F_{mj}c_j&\text{as $\left\{b_i,b_m^\dagger\right\}=\delta_{im}$}&\\&=b_m^\dagger f-F_{mj}c_j\implies\left[f,b_m^\dagger\right]=-F_{mj}c_j\end{align}$$
Similarly, $\left[c_m,g\right]=-G_{jm}b_j^\dagger=-G_{mj}b_j^\dagger$ (remembering $\boldsymbol G$ is symmetric).
Now consider:
$$\begin{align}f^nb_m^\dagger&=f^{n-1}\left(b_m^\dagger f-F_{mj}c_j\right)\\&=f^{n-2}\left(\left(b_m^\dagger f-F_{mj}c_j\right)f-F_{mj}c_j\right)\\&\enspace\,\quad\vdots\quad\enspace\,\,\text{recursively applying $\left[f,b_m^\dagger\right]=-F_{mj}c_j$}\\&=b_m^\dagger f^n-nf^{n-1}F_{mj}c_j\implies\left[f^n,b_m^\dagger\right]=-nf^{n-1}F_{mj}c_j\end{align}$$
This can be proved more formally using proof by induction; however, proof by induction requires one to guess the final form and the above method is how one would go about guessing that form. Similar arguments can be applied to show that $\left[c_m,g^n\right]=-nG_{mj}b_j^\dagger g^{n-1}$.
Now using these commutators we can prove the desired result as follows:
$$\begin{align}\langle\psi|b_m^\dagger&\equiv\langle0|e^fe^{\lambda g}b_m^\dagger&&\\&=\langle0|e^fb_m^\dagger e^{\lambda g}&\text{as $\left[g,b_m^\dagger\right]=0$}&\\&\equiv\langle0|\sum_{n=0}^\infty\frac{1}{n!}f^nb_m^\dagger e^{\lambda g}&\small{\begin{matrix}\text{as functions of operators are}\\\text{defined by their taylor series}\end{matrix}}&\\&=\langle0|\sum_{n=0}^\infty\frac{1}{n!}\left(b_m^\dagger f^{n}-nf^{n-1}F_{mi}c_i\right) e^{\lambda g}&\text{as $\left[f^n,b_m^\dagger\right]=-nf^{n-1}F_{mi}c_i$}&\\&=-\langle0|\sum_{n=0}^\infty\frac{1}{\left(n-1\right)!}f^{n-1}F_{mi}c_ie^{\lambda g}&\text{as $\langle0|b_m^\dagger=0$}&\\&\equiv-\langle0|e^fF_{mi}c_ie^{\lambda g}&&\\&\equiv-\langle0|e^fF_{mi}c_i\sum_{n=0}^\infty\frac{\lambda^n}{n!}g^n&&\\&=-\langle0|e^fF_{mi}\sum_{n=0}^\infty\frac{\lambda^n}{n!}\left(g^nc_i-nG_{ij}b_j^\dagger g^{n-1}\right)&\text{as $\left[c_i,g^n\right]=-nG_{ij}b_j^\dagger g^{n-1}$}&\\&\equiv-\langle0|e^fe^{\lambda g}F_{mi}\left(c_i-\lambda G_{ij}b_j^\dagger \right)&\text{as $\left[g,b_m^\dagger\right]=0$}&\\&\equiv-\langle\psi|F_{mi}\left(c_i-\lambda G_{ij}b_j^\dagger \right)&&\\\implies-\langle\psi|F_{mi}c_i&=\left(\delta_{mj}-\lambda F_{mi}G_{ij}\right)\langle\psi|b_j^\dagger&&\\\Longleftrightarrow-\langle\psi|\boldsymbol F\vec c&=\left(\boldsymbol{1}-\lambda \boldsymbol F\boldsymbol G\right)\langle\psi|\vec b^\dagger&\small{\begin{matrix}\text{in matrix notation: where}\\\text{matrices are bold and}\\\text{vectors have over arrows}\end{matrix}}&\\\implies\langle\psi|\vec b^\dagger&=-\langle\psi|\left(\boldsymbol{1}-\lambda \boldsymbol F\boldsymbol G\right)^{-1}\boldsymbol F\vec c\\&=-\langle\psi|\left(\boldsymbol F^{-1}-\lambda\boldsymbol G\right)^{-1}\vec c&\text{as $\left(\boldsymbol A\boldsymbol B\right)^{-1}=\boldsymbol B^{-1}\boldsymbol A^{-1}$}&\end{align}$$

Additionally, I just wanted to point out that while:
$$\langle0|e^{bFc}=\langle0|e^{\sum_{ij}b_i F_{ij} c_j}=\langle0|\prod_{ij}(1+b_iF_{ij}c_j)$$
is true it does not hold for a general bra $\langle\phi|$:
$$\langle\phi|e^{bFc}=\langle\phi|e^{\sum_{ij}b_i F_{ij} c_j}=\langle\phi|\prod_{ij}(1+b_iF_{ij}c_j+\ldots)\ne\langle\phi|\prod_{ij}(1+b_iF_{ij}c_j)$$
As functions of operators can be defined by their Taylor expansions and so the last expression has truncated the operator to first order in $b_iF_{ij}c_j$.
