The answer is yes indeed. It is a special case of brn function. $$ R=\frac{x^B-1}{x^N}=f_{B,N}(x) $$ $$ x=arcf_{B,N}(R)=brn_{B,N}(R) $$ $$ brn_{B,N}(R)=\sum_{g=0}^∞(\frac{R^g}{B^gg!}\prod_{r=1}^{g-1}(-Br+1+Ng)) $$ radius of convergence $$ \left|\frac{N^N(B-N)^{B-N}R^B}{B^B}\right|<1 $$ $$ B∈ℂ, N∈ℂ, R∈ℂ $$ a function named after the mathematician Bring.

article about ultraexponentiation and ultraroot.

calculator with brn button.

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.