$\def\srt{\operatorname{srt}}$Introduction:
There is a multiple series expansion for the super root $\srt_n(z)$ valid near $0.7<|z|<1.4$. However, for around $0<|z|<1.3$, there is this expansion:
$$\sqrt[k]z_s=\srt_k(z)=z-\sum_{n=1}^\infty\frac1{n!}\frac{d^{n-1}}{dt^{n-1}}e^{(n+1)t-e^t-\overbrace{ne^{t-e^{\dots}}}^{k-2\ “e^t”\text s}}\bigg|_{\ln(-\ln(z))}$$
Unfortunately, there is no obvious way to expand it as $k-1$ sums preserving convergence. A starting point is a method for finding a single series expansion for $\frac{d^{n-1}}{dt^{n-1}}e^{(n+1)(t-e^t)}$, when $k=3$, or the derivatives in: $$\ln(-\ln(\srt_3(e^z)))=\ln(-z)+\sum_{n=1}^\infty\frac1{n!}\left.\frac{d^{n-1}}{dt^{n-1}}e^{n(t-e^t)}\right|_{\ln(-z)}$$
converging around $0<|z|<1.3$
Attempt 1:
Trying $\srt_3(z)$ via $e^y$ Maclaurin expansion and changing summation order gives:
$$\srt_3(z)=1-\sum_{n=1}^\infty\sum_{m=0}^n\frac{(-1)^m m^{n-m} n^{m-2}}{(m-1)!(n-m)!}\ln^n(z)$$
converges for about $0.4<|z|<1.3$. If $n\gg m$, the region of convergence is a bit larger, but not the original $0<|z|<1.3$:
Attempt 2:
Using Stirling S2, general Leibniz rule, and factorial power $a^{(b)}$:
$$\frac{d^{n-1}}{dt^{n-1}}e^{n(t-e^t)}=\sum_{k=0}^{n-1}\sum_{m=0}^n s_{n-1}^{(k)}e^{-ne^t}e^{(k-m+n)t}\binom k nn^{(m)} (-n)^{k-m}$$
but this is not a single series expansion for:
$$\frac{d^n}{dt^n}e^{a(t-e^t)} =ae^{a(t- e^t)}\sum_{m=0}^{n-1} (-1)^m a_m e^{am}$$
Pattern:
Creating a table and using the OEIS shows that:
$$\frac{d^{n-1}}{dt^{n-1}}e^{(n+1)(t-e^t)}=(-1)^{n+1} ne^{(n+1)(t-e^t)}\left((-1)^{n+1} (n+1)^{n-2}+(-1)^n ((n+2)^{n-1}-(n+1)^{n-1})e^t+?e^{2t}+\dots+(n+1)^{n-2} e^{n-1}\right)$$
and
$$\frac{d^n}{dt^n}e^{t-e^t}=(-1)^n\left((-1)^n+(-1)^{n+1}S_{n+1}^{(2)}e^t+(-1)^nS_{n+1}^{(3)}e^{2t}+(-1)^{n+1}S_{n+1}^{(4)}e^{3t}+\dots+e^{nt}\right)$$ Question: What are methods for finding $\frac{d^n}{dt^n}e^{a(t-e^t)}$ as a single series so that a series expansion of $\srt_3(z)$ converges around $0<|z|<1.3$?
