Let $f : \Bbb Z^d \to \Bbb R$. The discrete average value $\overline{f(x)} : \Bbb Z^d \to \Bbb R$ is defined by $$\overline{f(x)}:=\frac{1}{2d}\sum_{y \in S_x}f(y)$$ where $S_x$ are the neighbors of $x$ in $\Bbb Z^d$. We define the discrete Laplacian $\Delta f:\Bbb Z^d \to \Bbb R$ as follows : $$ \Delta f(x):=\overline{f(x)}-f(x).$$ Let $D:=\{x_1,\dots,x_n\} \subset \Bbb Z^d$. We denote by $\Delta_D F$, the function that is equal to $\Delta F$ in $D$ and $0$ outside of $D$.
Now, let $(X_n)_{n≥0}$ be a simple random walk in $\Bbb Z^d$, with law denoted by $P_x$ when it is started at $x$. Let $τ = τ_D := \inf\{n ≥ 0 : X_n \notin D\}$ be its first exit time from $D$.
We define the Green’s function $G_D$ in $D$ to be the function defined on $D × D$ by $$G_D(x,y)=\Bbb E_x\bigg(\sum_{k=0}^{\tau-1}\mathbb 1_{\{X_k=y\}} \bigg) $$ $$ =\sum_{k=0}\#\{\text{paths }x \to y \text{ in } k \text{ steps within } D\}\times\big(\frac{1}{2d}\big)^k$$
The Green’s function $G_D$ is a symmetric function defined on $D × D$, so that it can be also written as a square symmetric matrix $(G_D(x_i,x_j))_{i,j≤n}$. (It can be shown that $G_D^{-1}=-\Delta_D$.)
I already have a proof involving Gaussian Free Field (which I don't know about) but what I want to do is to give a proof of the fact that $$\det G_D =G_D(x_1,x_1)×\det G_{D\backslash \{x_1\}} $$ only by using the fact that $$G_D(x,y) =\sum_{k=0}\#\{\text{paths }x \to y \text{ in } k \text{ steps within } D\}\times\big(\frac{1}{2d}\big)^k$$ Since $G_D$ is a symmetric matrix, I tried to get its eigenvalues but I feel its clearly a wrong way. I also thought of doing it very formally : since the determinant is a sum of product of entries and $G_D(x_1,x_1)$ is also a sum, I tried to show that the terms were equal by hand but it was overly complicated. Last attempt was some kind of induction argument (cf proofs about adjacency matrix in graph theory) but didn't lead me anywhere since $D$ is finite. Is there a way out ?