Consider in $\mathbb{R}^{n+1}$ with standard basis $e_1, \dots , e_{n+1}$ the following hyperplanes arrangement $$ \mathcal{H}=\{H_i | H_i=e_i^\perp \}$$
Now let consider the restiction $\mathcal{H}^0$ of this arrengement to the hyperplane $H^0=\{(x_1, \dots , x_{n+1}) | \sum x_i = 0\}$ (i.e. the hyperplanes of $\mathcal{H}^0$ are the set of the intersections $H_i \cap H^0 \subset H^0 \simeq \mathbb{R}^n$)
Which is the Characteristic polynomial of $\mathcal{H}^0$?
Remark: I use the index set $\{0,1,2,\ldots,n\}$ instead of $\{1,2,\ldots,n+1\}$.
Let $[n]:=\{0,1,2,\ldots,n\}$. For $I\subseteq [n]$, define $$\pi_I:=\text{span}_\mathbb{R}\big\{e_j\,|\,j\in[n]\setminus I\big\}=\bigcap_{i\in I}\,H_i\text{ and }\pi_I^0:=\pi_I\cap H^0\,.$$ Then, the intersection poset $P:=L\left(\mathcal{H}^0\right)$ contains all $\pi_I^0$ with $|I|<n$. Elements of a given rank $r$ of $P$ are of the form $\pi_I^0$ with $|I|=r$. Furthermore, the partial order $\preceq$ on $P$ is given by setting $\pi_I^0 \preceq \pi_J^0$ if and only if $I\subseteq J$ or $|J|=n$. Write $\pi^0$ for the point $(0,0,\ldots,0)=\pi^0_{I}$ for any $I\subseteq [n]$ with $|I|\geq n$.
Let $\mu:P\times P\to \mathbb{Z}$ be the Möbius function of the poset $P$. Observe that, for a given $I\subseteq [n]$ with $|I|<n$, we must have $$\mu\left(\pi_J^0,\pi_I^0\right)=(-1)^{|I|-|J|}$$ for all $J\subseteq I$. Furthermore, $\mu\left(\pi^0,\pi^0\right)=1$ and, for $I\subseteq [n]$ such that $|I|<n$, $$\mu\left(\pi_I^0,\pi^0\right)=-\sum_{s=0}^{n-1-|I|}\,(-1)^s\,\binom{n+1-|I|}{s}=(-1)^{n-|I|}\big(n-|I|\big)\,.\tag{*}$$ Thus, the characteristic polynomial of $\mathcal{H}^0$ is given by $$\chi_{\mathcal{H}^0}(t)=\sum_{v\in P}\,\mu\left(\pi_\emptyset^0,v\right)\,t^{\dim_\mathbb{R}(v)}=(-1)^n\,n+\sum_{d=1}^{n}\,(-1)^{n-d}\,\binom{n+1}{d+1}\,t^d\,.$$ That is, $$\chi_{\mathcal{H}^0}(t)=\frac{(t-1)^{n+1}-(-1)^n(t-1)}{t}\,.$$