Let $A$ be a hyperplane arrangement in $\mathbb{R}^n$ and $X\in A$. Define the restriction $A^X=\{H\cap X: H\in A, X\not\subseteq H \}$ and let $\mathcal{L}_{A^X}$ denote the corresponding geometric lattices. Show that $\sum_{X\in\mathcal{L}_A}\chi(A^X,q)=q^n$.
Attempt: Consider $\mathcal{L}_{A^X}$ as the sublattice $[X,\hat{1}]$ where $\hat{1}$ is the maximal element of $\mathcal{L}_A$. I wrote so far $\sum_{X\in\mathcal{L}_A}\chi(A^X,q)=\sum_{x\in\mathcal{L}_A}\sum_{z\in\mathcal{L}_{A^X}}\mu_{\mathcal{L}_{A^X}}(\hat{0},z)q^{dim(z)}=\sum_{x\in\mathcal{L}_A}\sum_{z\in [x,\hat{1}]}\mu_{\mathcal{L}_A}(x,z)q^{dim(z)}$
but I am not sure how to procceed or if I am on the right track.
Edit: I think I got it $\sum_{x\in\mathcal{L}_A}\sum_{z\in [x,\hat{1}]}\mu_{\mathcal{L}_A}(x,z)q^{dim(z)}=\sum_{x\in\mathcal{L}_A}q^{dim(x)}\sum_{z\in[\hat{0},x]}\mu_{\mathcal{L}_A}(z,x)$ and if $x\ne\hat{0}$ I get $\sum_{z\in[\hat{0},x]}\mu_{\mathcal{L}_A}(z,x)=0$ as it is the sum of the Mobius function over a closed ideal poset.