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 $f(x) = \frac{x}{1-e^{-x}}$.

The following is basically taking a standard proof of Lagrange inversion and specializing it to your case, but it might amuse. You can rewrite $(\star)$ as $$\frac{1}{2 \pi i} \oint \left(\frac{f(x)}{x}\right)^{n+1} dx =1$$ for all $n$, where the contour surrounds $0$ and is small enough to avoid all other poles of $f$. Multiplying by $y^{n+1}$ and summing on $n$, $$\frac{1}{2 \pi i} \oint \sum_{n=0}^{\infty} \left(\frac{y f(x)}{x}\right)^{n+1} dx =\frac{y}{1-y}$$ or $$\frac{1}{2 \pi i} \oint \frac{dx}{1-y f(x)/x} = \frac{y}{1-y}.$$ Set $g(x)=x/f(x)$. By the holomorphic inverse function theorem, $g$ is invertible near zero, set $h=g^{-1}$.

The only pole of the integrand near $0$ is at $x=h(y)$. The residue at that pole is $$\frac{-1}{y \frac{d}{du} g(u)^{-1}} = \frac{1}{y g(x)^{-2} g'(x)} = \frac{h'(y)}{y y^{-2}} = y h'(y).$$

So $y h'(y) = \frac{y}{1-y}$, $h'(y) = \frac{1}{1-y}$, $h(y) = -\log(1-y)$ (no constant of integration since $h(0)=0$), $g(x) = 1-e^{-x}$ and $f(x) = x/(1-e^{-x})$.

A pure topological derivation of the characteristic power series of the Todd class can be obtained by looking at pushforward maps in complex oriented cohomology theories. This is an outline of the argument: basically, since both topological complex $K$-theory and even $2$-periodic rational singular cohomology are complex oriented cohomology theories, it is possible to define integration maps (for stably complex $X$):

$$ \int_X^{K}:K(X) \to K(\mathrm{pt}) \cong \mathbb{Z} $$ $$ \int_X^{HP_{\mathrm{ev}}\mathbb{Q}}: \prod_{i\in \mathbb{Z}}H^{2i}(X;\mathbb{Q}) \to \prod_{i\in \mathbb{Z}}H^{2i}(\mathrm{pt};\mathbb{Q}) \cong \mathbb{Q}. $$ It is easy to see, by doing some consideration on how integration maps are defined, that these integration maps are non natural with respect to the Chern character $\mathrm{ch}$, and that there exists a class $\mathrm{td}_X \in \prod_{i\in \mathbb{Z}}H^{2i}(X;\mathbb{Q})$ (which is defined to be the Todd class of $X$) such that $\require{AMScd}$ \begin{CD} K(X) @>{\mathrm{ch}_X(-)\cdot \mathrm{td}_X}>> \prod_{i \in \mathbb{Z}}H^{2i}(X;\mathbb Q)\\ @V{\int_X^K}VV @VV{\int_X^{HP_{\mathrm{ev}}\mathbb{Q}}}V \\ \mathbb{Z} @>{\mathrm{ch}_{\mathrm{pt}}}>> \mathbb{Q} \end{CD} is commutative. The integration maps are defined by means of the Thom isomorphisms (multiplications by the Thom class) for complex vector bundles in a complex oriented cohomology theory. Now, by using the properties of these Thom classes and an application of the splitting principle, it is possible to show that the Todd class $\mathrm{td}_X$ can be expressed as a product of pullbacks of single cohomology class, namely $$ \dfrac{c_1(\mathcal{O}(1))}{1-e^{-c_1(\mathcal{O}(1))}} \in \prod_{i \in \mathbb{Z}}H^{2i}(\mathbb{P}^\infty;\mathbb{Q}) \cong \mathbb{Q}[[c_1(\mathcal{O}(1))]], $$ which shows that the characteristic power series of the Todd class is exactly $f(t)=t/(1-e^{-t})$.