Your equation can be written as an equation for exponential generating functions: $f(x) = g(x)(f(x)+1)$, where $f(x) = \sum_{n\ge1}n^nx^n/n!$ and $g(x) = \sum_{n\ge1}n^{n-1}x^n/n!$.

We can see that for those $f(x)$ and $g(x)$, we have $f(x) = xg'(x)$. If we then solve the differential equation $xg'(x) = \frac{g(x)}{1-g(x)}$ with $g(0)=0$, we get that the solution satisfies $x=g(x)e^{-g(x)}$.

By the Lagrange inversion formula, the computational inverse of $xe^{-x}$ is exactly our $g(x)$.

I'm sure some permutation of the reasoning steps above gives a proof for your equation.

Your equation can be written as an equation for exponential generating functions: $f(x) = g(x)(f(x)+1)$, where $$f(x) = \sum_{n\ge1}n^nx^n/n!$$ and $$g(x) = \sum_{n\ge1}n^{n-1}x^n/n!$$

We can see that for those $f(x)$ and $g(x)$, we have $f(x) = xg'(x)$. If we then solve the differential equation $$xg'(x) = \frac{g(x)}{1-g(x)}$$ with $g(0)=0$, we get that the solution satisfies $x=g(x)e^{-g(x)}$.

By the Lagrange inversion formula, the computational inverse of $xe^{-x}$ is exactly our $g(x)$.

I'm sure some permutation of the reasoning steps above gives a proof for your equation.