I am working through this book and I am stuck with the following: Let $\mathfrak{t}^N(\mathbb{R}^d)$ be the set of all truncated tensors in $T^N(\mathbb{R}^d):= \oplus_{k=0}^N(\mathbb{R}^d)^{\otimes k}$ with zeroth component being equal to zero and $1+\mathfrak{t}^N(\mathbb{R}^d)$ the set of truncated tensors with zeroth component being equal to 1. The authors introduce the exponential and logarithm like this: $$ \exp: \mathfrak{t}^N(\mathbb{R}^d) \to 1 + \mathfrak{t}^N(\mathbb{R}^d); a\mapsto 1+\sum_{k=1}^N \frac{a^{\otimes k}}{k!} \\ \log: 1 + \mathfrak{t}^N(\mathbb{R}^d) \to \mathfrak{t}^N(\mathbb{R}^d); 1+a\mapsto \sum_{k=1}^N (-1)^{k+1}\frac{a^{\otimes k}}{k}. $$ The authors claim that by a direct calculation it follows that $$ \exp(\log(1+a)) = 1 + a\quad \text{and} \quad \log\exp(a) = a $$ for $a \in \mathfrak{t}^N(\mathbb{R}^d)$.
Question: How can one actually prove this?
Attempt: I calculated this for $N \leq 3$ by hand, and from the usual series definitions of $\exp$ and $\log$ in $\mathbb{R}$ it seems true that the functions are each others inverse. However, I am stuck when showing this for general $N$. I calculate $$ \log(\exp(a)) = \sum_{k=1}^N \frac{(-1)^{k+1}}k\left(\sum_{l=1}^N \frac{a^{\otimes l}}{l!}\right)^{\otimes k} = \sum_{k=1}^N \frac{(-1)^{k+1}}k\left(\sum_{j_1+\dots+j_N=k}\frac{k!}{j_1!\dots j_N!}\left(\frac {a^{\otimes 1}}{1!}\right)^{\otimes j_1} \dots \left(\frac {a^{\otimes N}}{N!}\right)^{\otimes j_N}\right) $$ One can simplify the right hand side a little further, but it got me nowhere. Does this approach lead to the correct answer after some non-obvious sum manipulations I am not seeing or is there a better approach?
I would appreciate any help or hint on how to approach this problem.
Thanks in advance!
Edit: One approach I found (not writing down all the details and hopefully correct) sets $c_t = \exp(\log(1+at))$ and hence $$ \frac d{dt}c_t = \left(\frac{\partial}{\partial a}\exp(a)\right)\big|_{a = \log(1+at)} \frac{d}{dt}\log(1+at) = \exp(\log(1+at)) \frac{a}{1+ta} = c_t \frac{a}{1+ta} $$ But this ODE has the unique solution $c_t = 1+at$. Hence for $t = 1$, we have $\exp(\log(1+a)) = 1+a$, as we wanted. The calculation for $\log(\exp(a))$ is similar.
But I still wonder if there is some approach showing it by a direct calculation.
