Is it possible to prove that
$$\sum_{n \ge 0} (n+1)^{n-1}\frac{x^n}{n!} =(1- \sum_{n \ge 1} (n-1)^{n-1} \frac{x^n}{n!}) ^{-1}$$
Where $(n-1)^{n-1} := 1$ for $n=1$
Idea:use the Lagrange inversion theorem
$$\sum_{n \ge 0} (n+1)^{n-1}\frac{x^n}{n!} = \frac{1}{x}\sum_{n \ge1} {r(n) \frac{x^n}{n!}}= R(x) /x $$