I was reading Jack D'Aurizio's website matemate.it and learnt that
$$\sum_{k\geq1}\frac{k^{k}}{k!(4e)^{k/2}}=1\tag{1}$$
I worked out some other relatable values:
$$\sum_{k\geq1}\frac{k^{k+1}}{k!(4e)^{k/2}}=4$$
$$\sum_{k\geq1}\frac{k^{k+2}}{k!(4e)^{k/2}}=32$$
$$\sum_{k\geq1}\frac{k^{k+3}}{k!(4e)^{k/2}}=416$$
$$\sum_{k\geq1}\frac{k^{k+4}}{k!(4e)^{k/2}}=7552$$
$$\sum_{k\geq1}\frac{k^{k+5}}{k!(4e)^{k/2}}=176128$$
As I typed the number $176128$ into the search bar of OEIS, I recognized that these values coincides with [A005172]number of labeled rooted trees of subsets of an n-set :
Now it seems like
$$\sum_{k\geq1}\frac{k^{k+n}}{k!(4e)^{k/2}}=a(n+1)\tag{2}$$
I read over the linked OEIS webpage and saw that $a(n)$ is related to Stirling formula and Lambert W-function. How should I prove result (2)? and result (1)?
Process:
With the help of Wolfram Alpha, I recognized that $$\sum_{k\geq1}\frac{k^{k}}{k!(4e)^{k/2}}=\frac{-W\left(-\frac{1}{2\sqrt{e}}\right)}{W\left(-\frac{1}{2\sqrt{e}}\right)+1}$$ where $W$ is Lambert $W$-function. Then, we can use the result that $$W\left(-\frac{1}{2\sqrt{e}}\right)=-\frac{1}{2}$$ to prove that $$\sum_{k\geq1}\frac{k^{k}}{k!(4e)^{k/2}}=1$$ I am trying to work backward. I want to know how to transform the series into Lambert-$W$ function representation.