Trefethen and Bau, Numerical Linear Algebra, p. 208 states that Rayleigh quotient iteration (combining Rayleigh quotient estimate for eigenvalues and inverse power iteration) converges cubically
...the convergence is ultimately cubic in the sense that if $\lambda_J$ is an eigenvalue of $A$ and $v^{(0)}$ is sufficiently close to the eigenvector $q_J$, then $$\|v^{(k+1)} - (\pm q_J)\| = O(\|v^{(k)} - (\pm q_J)\|^3)$$ and $$|\lambda^{(k+1)} - \lambda_J| = > O(|\lambda^{(k)} - \lambda_J|^3)$$ as $k \rightarrow \infty$.
Their argument is as follows. Suppose that convergence occurs, and that $\|v^{(k)} - q_J\| \leq \epsilon$ for some small $\epsilon$. Then the Rayleigh quotient estimation for an eigenvalue gives an eigenvalue estimate $\lambda^{(k)}$ with $|\lambda^{(k)} - \lambda_J| = O(\epsilon^2)$.
Then apply the proof of the inverse power iteration for one step to obtain $v^{(k+1)}$ from $v^{(k)}$ and $\lambda^{(k)}$, so that $$\|v^{(k+1)} - q_J \|= O(|\lambda^{(k)}-\lambda_J| \|v^{(k)} - q_J\|) = O(\epsilon^3)$$ with the constants in the big-Oh symbols uniform in sufficiently small neighborhoods.
I don't see the left most equality above, nor why the constants are uniform.
It seems that all power iteration gives is $$O(|\lambda^{(k)}-\lambda_J|/|\lambda^{(k)}-\lambda_K|)$$ where $\lambda_K$ is the second closest eigenvalue.