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The power series for the inverse function of $ye^{-y}$
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\\
Noam D. Elkies
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For small $x$ let $y(x)$ be the small solution of $ye^{-y} = x$.
Abel showed that this function has the curious power series
$$
y = \sum_{n=1}^\infty n^{n-1} \frac{x^n}{n!}.
$$
One standard application of this is the enumeration of
trees on $n$ labeled vertices (there are $n^{n-2}$).
Abel's power series 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 $ye^{-y} = x$ to find
$$
1 = (1-y) e^{-y} y' = (1-y) \frac{x}{y} y'.
$$
Multiply by $y^\beta$ to obtain
$$
y^\beta = x (y^{\beta-1} - y^\beta) y'
= x \frac{d}{dx} \left(
  \frac{y^\beta}{\beta} - \frac{y^{\beta+1}}{\beta+1}
  \right)
\eqno{(1)}
$$
(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 (1) amounts to
$$
a_{m+\beta}(\beta) = (m+\beta)
\left(
 \frac{a_{m+\beta}(\beta)}{\beta} - \frac{a_{m+\beta}(1+\beta)}{1+\beta}
\right) \, ,
$$
which simplifies to the recursion
$$
a_{m+\beta}(\beta) =
  \frac{\beta}{1+\beta} \, \frac{m+\beta}{m} \,
a_{m+\beta}(1+\beta).
$$
Applying this $m$ times, we find
$$
a_{m+\beta}(\beta) =
  \frac{\beta}{m+\beta} \, \frac{(m+\beta)^m}{m!} \,
a_{m+\beta}(m+\beta).
$$
But $a_{m+\beta}(m+\beta)$ is the leading coefficient of $y^{m+\beta}$,
and thus equals~$1$.  We conclude that
$$
a_{m+\beta}(\beta) = \frac{\beta (m+\beta)^{m-1}}{m!} \, .
$$
In particular, taking $\beta=1$ and $m=n-1$
we find that the $x^n$ coefficient of~$y$ is $n^{n-1}/n!$, as claimed.
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