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Yet another proof of the series in~$x$ for the powers $y^\beta$
\\
of the solution of $y(1-y)^t = x$
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\vspace*{2ex}
Noam D. Elkies
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{\bf Proposition.}
{\em
Fix $t$, and for each $\beta \notin \{0, -1, -2, -3, \ldots\}$
let $S_{t,\beta}(x)$ be the power series
$$
S_{t,\beta}(x) := \sum_{j=0}^\infty
  \frac{\beta}{j+\beta} \, {t(j+\beta)+j-1 \choose j} \, x^{j+\beta}.
$$
Then $S_{t,\beta}(x) = y^\beta$, where $y = S_{t,1}(x)$
is the unique solution of $y(1-y)^t = x$ in $({\bf Q}(t))[[x]]$.
}

{\em Proof}\/: It is enough to check that
$S_{t,\beta}\bigl(y(1-y)^t\bigr) = y^\beta$.  We have
\begin{eqnarray*}
S_{t,\beta}\bigl(y(1-y)^t\bigr)
& = &
  \sum_{j=0}^\infty
  \frac{\beta}{j+\beta} \, {t(j+\beta)+j-1 \choose j} \,
  y^{j+\beta} (1-y)^{t(j+\beta)}
\\
& = &
  \sum_{j=0}^\infty
  \frac{\beta}{j+\beta} \, {t(j+\beta)+j-1 \choose j} \, y^{j+\beta} \,
  \sum_{k=0}^\infty (-1)^k {t(j+\beta) \choose k} y^t
\\
& = &
  y^\beta + \beta \sum_{m=1}^\infty y^{m+\beta}
  \sum_{j=0}^m \frac{(-1)^{m-j}}{j+\beta}
  \, {t(j+\beta)+j-1 \choose j} {t(j+\beta) \choose m-j}
  \; ,
\end{eqnarray*}
where we took $m=j+k$\/ in the last transformation.
Using the factorial formula for combinatorial coefficients
we rewrite this as
\begin{eqnarray*}
S_{t,\beta}\bigl(y(1-y)^t\bigr)
& = &
  y^\beta + \beta t \sum_{m=1}^\infty \frac{y^{m+\beta}}{m!} \,
  \sum_{j=0}^m (-1)^{m-j} {m \choose j} \,
    \frac
     {\bigl( t(j+\beta)+j-1 \bigr)!}
     {\bigl( t(j+\beta)+j-m \bigr)!} \; .
\end{eqnarray*}
But the inner sum is the $m$-th finite difference of
$$
\frac
 {\bigl( t(j+\beta)+j-1 \bigr)!}
 {\bigl( t(j+\beta)+j-m \bigr)!}
= \prod_{i=1}^{m-1} \bigl( t(j+\beta)+j-i \bigr),
$$
which as a function of $j$ is a polynomial of degree $m-1$.
Hence its $m$-th finite difference vanishes.  We conclude that
$S_{t,\beta}\bigl(y(1-y)^t\bigr) = y^\beta$, as desired.~~$\diamondsuit$

{\em Remark}\/: This approach was suggested by R.~Stanley's proof
of the power series for the solution of $y e^{-y} = x$
(see the bottom of page 28 of {\em Enumerative Combinatorics}, Vol.~2).
That proof can be recovered from ours by setting $\beta=1$
and taking a scaled limit as $k\rightarrow\infty$.

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