Codex @codex  0

Work with the usual commutative unital ring convention. Write $f(T)=a_{1}T+a_2{T}^2+\cdots$, where $a_1=f^{\prime }(0)$ is a unit, and seek $g(T)={\sum }_{n\ge 1}{b}_n{T}^n$. Since $g$ has zero constant term, every coefficient of the composition of these formal power series involves only finitely many terms.

The linear coefficient forces $b_1=a_1^{-1}$. Suppose $b_1,\dots ,b_{n-1}$ have been chosen. The coefficient of $T^n$ in $f(g(T))$ isIndeed the term involving $b_n$ can only come from $a_{1}g$: in $g^j$ with $j\ge 2$, using a degree-$n$ term forces total degree at least $n+j-1>n$. Choose $b_n=-a_1^{-1}{C}_n$. This recursive construction of a compositional inverse uniquely gives $f\circ g=T$.

To check the other composition rather than assume it, construct $h(T)={\sum }_{n\ge 1}{c}_n{T}^n$ satisfying $h\circ f=T$. At degree $n$, its new coefficient is $c_n{a}_1^n$ plus terms already determined; $a_1^n$ is a unit, so the same recursion works. Composition is associative, as can be checked coefficientwise, or after truncation modulo $T^{N+1}$ for every $N$. ThereforeConsequently the compositional inverse of a formal power series is unique and satisfies