In Prospects in Mathematics (AM-70), 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.

The only approach I see to reach this conclusion is:

- Use ($\star$) to find the first several terms: $b_1 = 1/2, b_2 = 1/12, b_3 = 0, b_4 = -1/720$.
- Notice that they look suspiciously like the coefficients in the exponential generating function for the Bernoulli numbers, so guess that $f(x) = \frac{x}{1-e^{-x}}$.
- 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 $f(x) = \frac{x}{1-e^{-x}}$.