let $\alpha >0$, and we define a sequence s.t $a_1 = \alpha $, and: $a_{n+1} = \sqrt{e^{a_n}-1}$, Prove that $a_n \rightarrow \infty$ 

let $\alpha >0$, and we define a sequence s.t $a_1 = \alpha $ 
and: $$ a_{n+1} = \sqrt{e^{a_n}-1}  $$ 
Prove that $a_n \rightarrow \infty$.


can I say that $e^x = 1 + x + \frac{x^2}{2!} + R_2(x) $ and: 
$e^{a_n} = 1 + a_n + \frac{a_n^2}{2!} + R_2(x) $
then : $e^{a_n} -1 > a_n + \frac{a_n^2}{2!}$
which means that $a_{n+1} > \sqrt{a_n + \frac{a_n^2}{2}}$
and then conclude from here that $a_n \rightarrow \infty$ ? 

First, note that $\frac{2\sqrt{6}}{6}> \frac{3}{5}$. (Any number between   $\frac{1}{2}$ and $\frac{2\sqrt{6}}{6}$ would do here.)
Next, for all $x>0$,
$$\frac{1}{x}+\frac{x}{6} \geq \frac{2\sqrt{6}}{6} > \frac{3}{5}$$ since that function of $x$ has its minimum at $x=\sqrt{6}$. 
Now look at the ratio $\frac{a_{n+1}^2}{a_n^2}$ where $a_n >0$:
$$
\frac{a_{n+1}^2}{a_n^2} = \frac{e^{a_n}-1}{a_n^2} > \frac{1+a_n + \frac{a_n^2}{2}+ \frac{a_n^3}{6}-1}{a_n^2}$$
$$=\frac{1}{2} + \left( \frac{1}{a_n} + \frac{a_n}{6} \right) > \frac{1}{2}+\frac{3}{5} = \frac{11}{10}
$$
So each $a_{n+1}^2 > \frac{11}{10}  a_n^2$.  So for all $n$
$$
a_{n} > \alpha \left(\sqrt{\frac{11}{10}}\right)^{n-1}$$ which clearly goes to infinity for any positive value of $\alpha$.
