Let $R$ be a $\mathbb{Q}-$algebra and let $D_0: R \to R$ be some non-trivial derivation on $R$. Moreover let $R[[T]]$ be formal power series ring. We define $D$ - extension of $D_0$ on $R[[T]]$ as $D\Big(\sum_{i=0}^\infty r_i T^i\Big) = \sum_{i=0}^\infty D_0(r_i)T^i.$
We define exponential of derivarion $\exp TD: R[[T]] \to R[[T]]$ as $ \exp TD(g) = \sum_{p\in\mathbb{N}_0} \frac{1}{p!}D^p(g)T^p$ for $g \in R[[T]]$. Now I want to show that $\exp TD$ is automorphism of ring $R[[T]].$ Showing that $\exp TD(f + g) = \exp TD(f) + \exp TD(g)$ is easy using addivity of $D_0$. I was also able to show that $\exp TD(fg) = \exp TD(f)\exp TD(g)$, using properties of Cauchy product. Also we can show that multiplication identity is mapped onto itself. So we can see that $\exp TD$ is endomorphism of $R[[T]]$.
I want to proof bijectivity of $\exp TD$ by showing that for every $f \in R[[T]]$ there exists $\exp TD^{-1}$. In fact I want to show that $\exp TD \circ \exp -TD = \exp -TD \circ \exp TD = $id$_{R[[T]]}$. With those calculations I have some problems. For examples how to show that $\sum_{p\in\mathbb{N}_0} \frac{1}{p!}D^p\Big(\sum_{q\in\mathbb{N}_0} \frac{1}{q!}D^q(f)(-T)^q\Big)T^p = \sum_{p\in\mathbb{N}_0} \frac{1}{p!}\Big(\sum_{q\in\mathbb{N}_0} \frac{1}{q!} D^{p+q}(f)(-T)^q \Big)T^p = f$? I wolud be grateful for any hints.
You are right that the inverse of $\exp TD$ is $\exp (-T)D$, which sends $g$ to $\sum_{p \in \Bbb N_0} \frac 1{p!} D^p(g)(-T)^p$.
To conclude your proof, you only need the following identity: $$\sum_{p + q = n} \frac 1{p!q!}(-1)^q = \begin{cases}1, &\textrm{if }n = 0;\\0, &\textrm{if }n > 0.\end{cases}$$ This identity becomes obvious when you multiply it by $n!$, which then becomes $\sum_{p + q = n}\binom n q (-1)^q = (1 - 1)^n$.