If $\alpha$ is a irrational number prove that $\operatorname{ord}\alpha \geq 1+\exp(\limsup \log(\log(a_n+1))/n)$
I tried the simple things like use the well known formula $\operatorname {ord}\alpha = 1+\limsup [\log q_{n+1}/\log q_n]$ and give lower and upper bounds to $q_n$ with the formula $q_{n+1}=a_nq_n+q_{n-1}$ (for example $q_{n+1} \leq \prod_{k=1}^n(a_k+1)$)
Remark1: Order of a real number is the greatest real number $v$ such that exist infinity rationals $p/q$, $p,q \in \mathbb{Z}$ that satisfy $|\alpha−p/q|≤1/q^v$
Remark2: $a_n$ are the coefficients of the continued fraction of $\alpha$ and $q_n$ is the sequence definied by the recurrence defined by the equation given and $q_0=1, q_1=1$