Show that $\alpha:=[2;2^{1!},2^{2!},2^{3!},..]$ (continued fraction) is transcendental Show that $\alpha:=[2;2^{1!},2^{2!},2^{3!},..]$ (continued fraction) is transcendental.
For clarification, $$\alpha = 2 + \cfrac{1}{2^{1!} + \cfrac{1}{2^{2!} + \cfrac{1}{2^{3!} + \cfrac{1}{\ddots}}}}$$

Here is my attempt: 
We know from Liouville's theorem that if $\alpha$ is $d-algebraic$, then $\exists_{0<C \in \Bbb R}\forall_{\frac{p}{q} \in \Bbb Q} |\alpha-\frac{p}{q}|>\frac{C}{q^d}$. 
Since $\alpha$'s continued fraction is infinite, then $\alpha \notin \Bbb Q$. So we can assume $2 \leq d$.
 We also know that for the convergents series the following holds: $|\alpha-\frac{p_n}{q_n}|<\frac{1}{a_{n+1}q^2}$. 
Putting it all together, we get for convergents series: $\frac{C}{q_n^d} < \frac{1}{a_{n+1}q_n^2}$. Hence $C < \frac{q_n^{d-2}}{a_{n+1}} = \frac{q_n^{d-2}}{2^{(n+1)!}}$.
Now I'm stuck. I would like to show that as $n \rightarrow \infty$, $C \rightarrow 0$. Which yields a contradiction for all $d$, proving that $\alpha$ is transcendental.
Any clues?

Apparently, you need an upper estimate for $q_n$. Since $q_n=a_nq_{n-1}+q_{n-2}\leqslant(a_n+1)q_{n-1}$, (note 1) we may conclude that
$$q_n\leqslant(a_1+1)\cdot{\dots}\cdot(a_n+1)\leqslant {\rm const}\cdot\prod_{i=1}^n a_i\qquad^{(\rm note\;2)}$$
and hence $q_n^2=o(a_{n+1})$. (note 3)



*

*My definitions may or may not be off-by-one relative to yours; that doesn't matter.

*To obtain this estimate, use the fact that the product $\prod\limits_{i=1}^\infty\left(1+{1\over a_i}\right)$ converges to a finite limit iff the sum $\sum\limits_{i=1}^\infty{1\over a_i}$ does the same, which it quite obviously does.

*Thanks to the Roth's theorem, even a much slower growth of $a_n$ would suffice to ensure the transcendental nature of $\alpha$.

