$N$ is approximately linear in $d$ for $N^d=\frac12 e^{N}$ let us look at the function $N^d e^{-N}$, for each $d\in \mathbb{N}$. The graphs of the function for various values of $d$ show a striking phenomenon: the graph look parallel, and with a near-constant distance between each consecutive graphs. 

Put in other words, Let $p<1$, and for each $d\in \mathbb{N}$ look at the $N$ that satisfies $N^de^{-N}=p$. Then it seems that the resulting function $N(d)$ is $o(n^2)$ or even $\Theta(n)$, and furthermore that  for all $d$, $N(d+1)-N(d)$ is approximately uniform in $p$.
This phenomenon is observed here without explaining the exact argument. This also gives a good statistical motivation for asking the question.
What will be a rigorous statement of this phenomenon, and why is it true?
 A: For all $d \geq e$ and $0 < p < 1$ the equation
$$
N^d = pe^N
$$
has two positive solutions $N$, call them $N_1$ and $N_2$. Since the maximum of $x \mapsto x^d e^{-x}$ occurs at $x=d$ and $x^d e^{-x} \to 0$ as $x \to \infty$ we know that $N_1 < d < N_2$.
We'll first investigate the behavior of $N_1$, then $N_2$ afterwards.
Because the maximum of $x \mapsto x^d e^{-x}$ is located at $x=d$, if $1 + \epsilon < x \leq d$ for some $\epsilon > 0$ then
$$
x^d e^{-x} > (1+\epsilon)^d e^{-1-\epsilon} \to \infty
$$
as $d \to \infty$. In this case the equation $x^d e^{-x} = p$ will not hold for $d$ large enough. Consequently we may conclude that
$$
\limsup_{d \to \infty} N_1 \leq 1.
$$
This is clearly not the solution we're looking for, so let's move on to $N_2$.
We know that $N_2 > d$. Taking logarithms of the original equation we have
$$
d \log N_2 = N_2 + \log p, \tag{$*$}
$$
and so
$$
N_2 > d \log d - \log p.
$$
In fact this captures the asymptotic behavior of $N_2$. Indeed, as $d \to \infty$ we know that $N_2 \to \infty$, so
$$
N_2 + \log p \sim N_2
$$
as $d \to \infty$, and thus equation $(*)$ implies
$$
d\log N_2 \sim N_2
$$
as $d \to \infty$. Taking logarithms again we see that
$$
\log d = \log N_2 - \log\log N_2 + o(1)
$$
as $d \to \infty$. Dividing both sides by $\log N_2$ yields
$$
\frac{\log d}{\log N_2} = 1 - \frac{\log\log N_2}{\log N_2} + o(1) = 1 + o(1)
$$
and thus $\log N_2 \sim \log d$. Plugging this back into $(*)$ allows us to conclude that
$$
N_2 \sim d\log d.
$$
