Nested logarithm sequence $a_{n+1}=a_n+\log a_n$, growth rate

It isn't a power law, and $\lim_{n\to\infty}r_n=1$. One way to prove this is to check by induction that $a_n\leq 2n\log n$ for every $n\geq 2$.

Hint. Note that if we show that eventually $$c_1 n\ln n\leq a_n \leq c_2 n\ln n\quad \text{with $0<c_1\leq c_2$}$$ ($c_1=1/2$ and $c_2=2$ should work by induction), then $\ln(a_n)\sim \ln(n)$, and, by Stolz-Cesaro Theorem, $$\lim_{n\to +\infty}\frac{a_n}{n\ln n}=\lim_{n\to +\infty}\frac{a_{n+1}-a_n}{(n+1)\ln (n+1)-n\ln(n)}= \lim_{n\to +\infty}\frac{\ln(a_n)}{\ln ((1+1/n)^n)+\ln(n+1)}=1.$$