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][1] about [ultraexponentiation][2] and ultraroot.

[calculator][3] with brn button.


  [1]: http://glax-plato.ru/exam/Math/brn.pdf
  [2]: https://www.cyberforum.ru/blog_attachment.php?attachmentid=8110&d=1687342320
  [3]: http://glax-plato.ru/