Solving the geometric series for q Is there a general way to find the $q > 0$ solving the equation from the geometric series $$1+q+q^2+q^3+\ldots + q^n = a$$ or $$\frac{1-q^{n+1}}{1-q} = a\quad\text{with } q \neq 1$$ for $a > 1$ and $n\in\mathbb N$?

My thoughts: Since polynomials aren't solvable in general for degree 5 or higher, I guess the above equation has no explicit solution for $n\ge 5$. In this case numerical approximations can be used. For $n=5$ also this method can be used.

Re-arrange 
\begin{align*}
  q &= \frac{a-1}{a-q^n} \\
    &= \frac{a-1}{a-\left(\frac{a-1}{a-q^n}\right)^n} \\
    &= \frac{a-1}
            {a-\left(
                 \frac{a-1}
                      {a-\left(
                           \frac{a-1}{a-\ddots}
                         \right)^n}
               \right)^n} \\
   &= \frac{a-1}{a}
      \left \{
        1+\frac{(a-1)^n}{a^{n+1}}+\ldots+
        \frac{(nk+k)!}{k! (nk+1)!}
        \left[ \frac{(a-1)^n}{a^{n+1}} \right]^{k}+\ldots
      \right \}
\end{align*}

Lagrange inversion formula
\begin{align*}
 f(q) &= aq-q^{n+1} \\
 q &= f^{-1}(a-1) \\
 &= \sum_{i=1}^{\infty} \frac{1}{i!} \frac{d^{i-1}}{dx^{i-1}} \left. \left(\frac{x-0}{f(x)-f(0)} \right)^{i} \right|_{x=0} [(a-1)-f(0)]^{i} \\
 &= \sum_{i=1}^{\infty} \frac{1}{i!} \frac{d^{i-1}}{dx^{i-1}} \left. \left(\frac{1}{a-x^{n}} \right)^{i} \right|_{x=0} (a-1)^{i}
\end{align*}

