The answer is yes indeed. It is a special case of $\DeclareMathOperator{\brn}{brn}\brn$ function.
$$
R=\frac{x^B-1}{x^N}=f_{B,N}(x)
$$
$$
x=\operatorname{arc}f_{B,N}(R)=\brn_{B,N}(R)
$$
$$
\brn_{B,N}(R)=\sum_{g=0}^\infty\left(\frac{R^g}{B^gg!}\prod_{r=1}^{g-1}(-Br+1+Ng)\right)
$$
radius of convergence
$$
\left|\frac{N^N(B-N)^{B-N}R^B}{B^B}\right|<1
$$
where $B, N, R \in C$.
The function $\brn$ is named after the mathematician Erland Samuel Bring.
Here there is an article about ultraexponentiation and ultraroot, while here there is a calculator with $\brn$ button.
