A limit resulting in the exponential generating function for labeled forests I found the following claim on OEIS and I am unsure how I would go about verifying it.
$$
\lim\limits_{N \rightarrow \infty} \left( \sum\limits_{n \geq 0}(N+n)^n \frac{x^n}{n!} \right)^{1/N} = \sum\limits_{n \geq 0} (n+1)^{n-1}\frac{x^n}{n!}
$$
From context in the entries, it seems that one way may be to make the substitution $N \mapsto y$, drop the limit, and take the Log of the left. Then if we send $y \rightarrow \infty$, the only terms that do not blow up are those with a $y^1$, so we just consider those and exponentiate to get the result.
Any ideas or references are appreciated!
 A: The species of labelled trees has the specification
$$\mathcal{T} =
\mathcal{Z} \times \mathfrak{P}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$
Now with Cayley's theorem saying that
$$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}$$
it is not difficult to see that
$$A(z) = \sum_{n\ge 0} (n+1)^{n-1} \frac{z^n}{n!}
= \frac{T(z)}{z}.$$
Observe that a labeled forest has specification
$$\mathfrak{P}(\mathcal{T})$$
which yields $\exp(T(z))$ and we once more have $A(z) = T(z)/z.$
Extracting coefficients via Lagrange inversion we find with $N\ge 1$
$$n! [z^n] A(z)^N = n! [z^n] \frac{T(z)^N}{z^N}
= n! [z^{N+n}] T(z)^N =
\frac{n!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{N+n+1}} T(z)^N \; dz.$$
Put $T(z)=w$ so that $z=w/\exp(w) = w\exp(-w)$ and
$dz = \exp(-w) - w\exp(-w)$
to get
$$\frac{n!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(w(N+n+1))}{w^{N+n+1}}
\times w^N\times (\exp(-w) - w\exp(-w)) \; dw
\\ = \frac{n!}{2\pi i}
\int_{|w|=\gamma}
\frac{\exp(w(N+n))}{w^{n+1}} (1 - w) \; dw.$$
But we have
$$n! [w^{n}] \exp(w (N+n)) = (N+n)^n$$
and
$$n! [w^{n-1}] \exp(w (N+n)) = n (N+n)^{n-1}.$$
This is
$$(N+n)^n - \frac{n}{N+n} (N+n)^n$$
andh therefore the  asymptotics of the coefficients in  $N$ are indeed
given by
$$\bbox[5px,border:2px solid #00A000]{
(N+n)^n \times \frac{N}{N+n}}$$
and with $n$ fixed $\lim_{N\rightarrow\infty} \frac{N}{N+n} = 1.$
The question is  why we would need  to take a limit  given that the
exact formula is so simple.

The labelled tree function recently appeared at this
MSE link.
