In [Prospects in Mathematics (AM-70)](http://books.google.com/books?id=zK7jmvsI_2wC&lpg=PP1&ots=LpN2oUA88m&dq=hirzebruch%20%22prospects%20in%20mathematics%22&pg=PA7#v=onepage&q&f=false), Hirzebruch gives a nice discussion of why the formal power series $f(x) = 1 + b\_1 x + b\_2 x^2 + \dots$ defining the Todd class must be what it is. In particular, the key relation $f(x)$ must satisfy is that 


($\star$) the coefficient of $x^n$ in $(f(x))^{n+1}$ is 1 for all $n$.


As Hirzebruch observes, there is only one power series with constant term 1 satisfying that requirement, namely
$$f(x) = \frac{x}{1-e^{-x}} = 1 + \frac{x}{2}+\sum\_{k\geq 2}{B\_{k}\frac{x^{k}}{k!}} = 1 + \frac{x}{2} + \frac{1}{6}\frac{x^2}{2} - \frac{1}{30}\frac{x^4}{24} + \dots,$$
where the $B\_k$ are the [Bernoulli numbers](http://en.wikipedia.org/wiki/Bernoulli_number). 


The only approach I see to reach this conclusion is: 


1. Use ($\star$) to find the first several terms: $b\_1 = 1/2, b\_2 = 1/12, b\_3 = 0, b\_4 = -1/720$.
2. Notice that they look suspiciously like the coefficients in the exponential generating function for the Bernoulli numbers, so guess that $f(x) = \frac{x}{1-e^{-x}}$.
3. Do a residue calculation to check that this guess does satisfy ($\star$).



> 
> My question is whether anyone knows of a less guess-and-check way to deduce from ($\star$) that $f(x) = \frac{x}{1-e^{-x}}$.
> 
> 
> 



Hirzebruch's motivation of the Todd class
Since you mention playing around with residues, I'm probably not telling you anything you don't already know. But there is a systematic way to extract the power series $f$ from
the coefficients of $x^{n-1}$ in $f(x)^{n}$, which goes by the name of the Lagrange inversion formula.


Assume that the constant term of $f$ is invertible, and define $g(x) = \frac{x}{f(x)}$.
Then $g(x)$ is a power series which has a compositional inverse. Denote this inverse by $h$, so that if $y = g(x)$ then $x = h(y)$. Write $h(y) = c\_1 y + c\_2 y^2 + c\_3 y^3 + \cdots$.
For every integer $n$, the product $n c\_n$ is the residue of the differential
$\frac{1}{y^n} h'(y) dy = \frac{1}{g(x)^{n}} dx = \frac{ f(x)^{n} }{x^n} dx$, which is the coefficient of $x^{n-1}$ in $f(x)^{n}$.


In your example, you get $c\_n = \frac{1}{n}$, so that $h(y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots = - \log(1-y)$. Then $g(x) = 1 - e^{-x}$, so that
$f(x) = \frac{x}{1-e^{-x}}$.


