TITLE: Show that the formal power series $ Q(x)=\frac{x}{1-e^{-x}}$ has the property that the coefficient of $x^n$ in $Q(x)^{n+1}$ is always $1$
QUESTION [7 upvotes]: When I am looking up the Wikipedia page for the definition of Todd class, it says that the formal power series defined by
$$
Q(x)=\frac{x}{1-e^{-x}}=1+\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\cdots$$
has the property that the coefficient of $x^n$ in $Q(x)^{n+1}$ is always $1$. It seems that this property does not immediately follow from the definition. How can we show that $Q(x)$ satisfies this property and if a power series satisfies this property, then it must be $Q(x)$?

REPLY [2 votes]: A way to circumvent "laborious" computations is to use Lagrange inversion theorem.
Applied to $z=f(w):=1-e^{-w}$ around $w=0$, it gives $w=\sum_{n=1}^\infty(g_n/n!)z^n$, where $$g_n=\frac{d^{n-1}}{dw^{n-1}}\left(\frac{w}{1-e^{-w}}\right)^n\Bigg|_{w=0}=(n-1)![w^{n-1}]Q(w)^n\qquad(n>0)$$ and, knowing that $-\log(1-z)=\sum_{n=1}^\infty z^n/n$, we get $[w^{n-1}]Q(w)^n=1$ for all $n>0$.