I'm trying to prove Exercise 1.7 from Rough Path Theory which states any $\xi=(1,\xi_1,\cdots,\xi_N) \in T^N(V)$ has an inverse and the following holds $$ \xi\, \otimes\sum_{n=0}^N (-1)^n(\xi-\mathbf{1})^{\otimes n}=\mathbf{1} \tag{1}$$ where is the unit $\mathbf{1}\triangleq(1,0,\cdots,0) \in T^N(V)$.
$T^N(V)$ is the $N$-th truncated tensor algebra over the real vectorspace $V$ and is defined by $$T^N(V) \triangleq \bigoplus_{n=0}^N V^{\otimes n}=\{\xi=(1,\xi_1,\cdots,\xi_N) : \xi_n \in V^{\otimes n} ,0\leq n\leq N\}$$ with the convention $V^{\otimes 0}\triangleq\mathbb{R}$.
The 1. question is how to prove $(1)$? My attempts will follow.
The 2. question is does $V^{\otimes 0}\triangleq\mathbb{R}$ imply $v^{\otimes 0}=\mathbf{1}$ for $v \in V$ especially for $v=0$? Only if assume so I'm able to prove it for $N=0,1,2,3$.
$\mathbf{My Attempts}:$
I calculated $(1)$ for $N=1,2,3$. Thereby I realized $$(\xi-\mathbf{1})^{\otimes n}=(\underbrace{0,\cdots,0}_{n \, times},\star,\cdots\star) \tag{2}$$ and can also proof this. From there on I can also proof that $$\Big(\xi\, \otimes\sum_{n=0}^N (-1)^n(\xi-\mathbf{1})^{\otimes n}\Big)_{l=1}=1.$$ What is left is that for $1<l\leq N$ $$\Big(\xi\, \otimes\sum_{n=0}^N (-1)^n(\xi-\mathbf{1})^{\otimes n}\Big)_{l>1}=0.$$ One way to prove it might be to figure out the $\star$ part of $(2)$ and continue via induction. I think it'll work but looks like a lot of work for this result what looks to me like a standard result. I tried the straightforward way and got \begin{align*} \Big(\xi\, \otimes\sum_{n=0}^N (-1)^n(\xi-\mathbf{1})^{\otimes n}\Big)_{l}&=\sum_{k=0}^l \xi_k \otimes \Big(\sum_{n=0}^N (-1)^n (\xi-\mathbf{1})^{\otimes n}\Big)_{l-k}\\ &=\sum_{k=0}^l \sum_{n=0}^N (-1)^n \xi_k \otimes \Big((\xi-\mathbf{1})^{\otimes n}\Big)_{l-k}\\ &=\sum_{k=0}^l \sum_{n=0}^{l-k} (-1)^n \xi_k \otimes \Big((\xi-\mathbf{1})^{\otimes n}\Big)_{l-k} \end{align*} where I'm stuck.