In trying to solve $x^{ax}=x+1$, like for the Foias Ewing constant equation $z^{z+1}=(z+1)^z$, one uses Lagrange reversion which seems to only converge in the following form after letting $x=e^t$:
$$x^{ax}=x+1\iff t=\frac1ae^{-t}\ln(e^t+1)\implies x=1+\sum_{n=1}^\infty\frac{a^{-n}}{n!}\left.\frac{d^{n-1}}{dt^{n-1}}e^{(1-n)t}\ln^n(e^t+1)\right|_0$$
One could easily expand $\ln(x+1)^n$ using the Stirling number of the first kind $S_n^{(m)}$ and find $\frac{d^{n-1}}{dt^{n-1}}e^{(1-n+m)t}\Big|_0$ to get a double sum solution, but $x=e^0=1$ is on the radius of convergence. Expanding $\ln^n(e^x+1)=\sum\limits_{k=n}^\infty(-1)^{n+k}\frac{n!}{k!}S_k^{(n)}(e^{-x}+1)^{-k}$ gives:
$$x=1+\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{a^{-n}}{k!}(-1)^{n+k}S_k^{(n)}\left.\frac{d^{n-1}}{dt^{n-1}}e^{(1-n)t}(e^{-t}+1)^{-k}\right|_0$$
The derivative almost looks like the Norlund polynomials . Also, Binomial series converges for $e^t<1$ or $e^t>1$, but not $e^t=1$. There is a way to use expand it using Stirling numbers of the second kind $s_n^{(m)}$, general Leibniz rule, and $\frac{d^n}{dx^n}\ln^a(x)$:
$$x=1+\ln(2)+\sum_{n=2}^\infty\sum_{k=1}^{n-1}\sum_{m=0}^k\sum_{j=0}^m\binom kms_{n-1}^{(k)}\frac1{n!}(1-n)^{(k-m)}(n-j+1)_jS_m^{(j)}2^{-m}\ln(2)^{n-j}$$
with the Pochhammer symbol $(x)_n$ and factorial power $x^{(n)}$. However, it is a quadruple series and expanding creates negative arguments for the gamma functions.
How does one evaluate $\displaystyle \left.\frac{d^{n-1}}{dt^{n-1}}e^{(1-n)t}\ln^n(e^t+1)\right|_0$ as a single or double sum?
