The region of convergence of the series $\displaystyle \sum_{n=1}^\infty \frac{1}{2^{3^n}}(z^2-z)^{3^n}$

Question

I was asked to find the region of convergence of the series $\displaystyle \sum_{n=1}^\infty \frac{1}{2^{3^n}}(z^2-z)^{3^n}$. I guess it is $1$ and my thought is that I first write it as \begin{equation*} \sum_{m=1}^\infty a_mz^m, \end{equation*} where $a_m=0$ for $3^n<m<3^{n+1}$. Then it is a gap series and I hope to find a subsequence $\{a_{m_k}\}$ such that \begin{equation} \lim_{k \to \infty} \sqrt[m_k]{a_{m_k}}=1, \end{equation} but I failed. Please give me some hints.