How to bound logarithm with variable base? (asymptotics) Problem: For any $n>1$ let
$\theta(n) = \big( 1 - (\ln n + 1)^{-1} \big)^{-1}$, where $\ln$ is the natural logarithm. Show that for some $k$ (not depending on $n$) it holds that
$\log_{\theta(n)}n \in \mathcal{O}((\log n)^k)$, i.e. there exists a positive real number $c$ such that $\log_{\theta(n)}n \leq c (\log n)^k$ for all large enough $n$ (where $\log$ is with respect to any base, e.g. 2).
What I know: k=1 is probably not good enough, but $k=2$ seems to be fine (looking at plots).
Preferably: $k$ as small as possible.
Motivation: If the claim is true, then I have an asymptotically fast algorithm for encoding certain algebraic geometry codes (error-correction).

Hint
$$y=\log_{\theta(n)}(n)=\frac{\log (n)}{\log (\theta (n))}$$
$$\theta (n)=\frac{1}{1-\frac{1}{\log (n)+1}}\implies y=\frac{\log (n)}{\log \left(\frac{1}{1-\frac{1}{\log (n)+1}}\right)}=-\frac {\log (n)}{\log\left({1-\frac{1}{\log (n)+1}}\right)}=-\frac{\log (n)}{\log \left(\frac{\log (n)}{\log (n)+1}\right)}$$
$$y=\frac{\log (n)}{\log \left(1+\frac{1}{\log (n)}\right)}$$
Make $n=e^x$, simplify and use Taylor series for large $x$.
