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\begin{center}
\begin{large}
The power series for the inverse function of $y(1-y)^t$
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\\
Noam D. Elkies
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Fix $t$\/ (usually an integer, but may be real or even complex);
For small $x$ let $y(x)$ be the small solution of $y(1-y)^t = x$.
Let $a_n$ be the coefficient of the power series expansion
$$
y = \sum_{n=1}^\infty a_n x^n.
$$
For $t=2$, the $a_n$ are Catalan numbers;
larger integers~$t$\/ yield generalized Catalan numbers.
The formula
$$
a_n = \frac1n {(t+1)n - 2 \choose n-1}
\eqno{(1)}
$$
is usually proved using the residue theorem.
The following derivation uses only elementary calculus,
and as an added bonus gives the power series expansion of
$y^\beta$ for all~$\beta$.

Differentiate both sides of $y(1-y)^t = x$ to find
$$
1 = \Bigl( 1 - \frac{ty}{1-y} \Bigr) (1-y)^t y'
  = \frac{1-(t+1)y}{1-y} \, \frac{x}{y} \, y'.
$$
Multiply by $(1-y) y^\beta$ to obtain
$$
y^\beta - y^{1+\beta} = x \bigl( y^{\beta-1} - (t+1) y^\beta \bigr) y'
= x \frac{d}{dx} \left(
  \frac{y^\beta}{\beta} - (t+1) \frac{y^{\beta+1}}{\beta+1}
  \right)
\eqno{(2)}
$$
(for $\beta \neq 0,-1$).
Now recall that the operator $x \, d/dx$ takes any power series
$\sum_n a_n x^n$ to $\sum_n n a_n x^n$.  Thus if for any~$c$
we expand $y^c$ in a power series
$$
y^c = \sum_{m=0}^\infty a_{m+c}(c) x^{m+c}
$$
then our identity (2) amounts to
$$
a_{m+\beta}(\beta) - a_{m+\beta}(1+\beta) = (m+\beta)
\left(
 \frac{a_{m+\beta}(\beta)}{\beta}
 - (t+1)\frac{a_{m+\beta}(1+\beta)}{1+\beta}
\right) \, ,
$$
which simplifies to the recursion
$$
a_{m+\beta}(\beta) =
  \frac{\beta}{1+\beta} \, \frac{t(m+\beta)+m-1}{m} \,
a_{m+\beta}(1+\beta).
$$
Applying this $m$ times, we find
$$
a_{m+\beta}(\beta) =
  \frac{\beta}{m+\beta} \, {t(m+\beta)+m-1 \choose m} \,
a_{m+\beta}(m+\beta).
$$
But $a_{m+\beta}(m+\beta)$ is the leading coefficient of $y^{m+\beta}$,
and thus equals~$1$.  We conclude that
$$
a_{m+\beta}(\beta) =
  \frac{\beta}{m+\beta} \, {t(m+\beta)+m-1 \choose m} \, .
$$
In particular, taking $\beta=1$ and $m=n-1$
we recover the formula (1) for the $x^n$ coefficient
$a_n = a_n(1)$ of~$y$.
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