Proving that a sequence diverges 
Prove that the sequence $a_{n+1}=a_n^2-2$ diverges if $|a_0|>2$.

What I did is proving that $a_{n+1}>a_n$.Then I defined $d_n=a_n-a_{n-1}$ and i've proved that the sequence $d_n$ is increasing too.
Is this enough to conclude the proof or i'm missing something?
Are there other proof to this fact?
 A: Yes, if you know the $d_n$ are positive and increasing that is enough to conclude the sequence diverges.  Explicitly, since $d_n\geq d_1$ for all $n$, you have $a_n=a_0+d_1+d_2+\dots+d_n\geq a_0+nd_1$, which grows without bound since $d_1$ is just some positive constant.  So more specifically, you can conclude that $a_n$ diverges to $\infty$.
(Note, however, that in this case it's not quite true that $d_n$ is increasing.  For instance, if $a_0=-2.1$ then $a_1=2.41$ and $a_2=3.8081$ so $d_1=4.51$ and $d_2=1.3981$.  What is true is that the sequence $(d_n)$ is eventually increasing, and a similar argument still shows then that $a_n$ must diverge to $\infty$.)
A: Also, if we take real constants $A,B > 0,$ with
$$ AB = 1 $$
$$ A + B = |a_0|,  $$ then at least one of $A,B$ is strictly larger than $1,$ and for $n \geq 1,$
$$ a_n = A^{\left(2^n \right)} + B^{\left(2^n\right)} $$ 
Demanding $A>B > 0$ we get
$$ A = \frac{|a_0| + \sqrt {a_0^2 -4}}{2}, $$
$$ B = \frac{|a_0| - \sqrt {a_0^2 -4}}{2}, $$
both real numbers because $a_0^2 > 4.$ Then $A > 1$ as $A+B > 2.$
A: To your question, "Are there other proof to this fact":
When $|a_0|>2$ you have for some $\epsilon >0$
$$a_1=a_0^2-2=2+\epsilon$$
$$a_2=a_1^2-2=(2+\epsilon)^2-2>2+4\epsilon$$
$$a_3=a_2^2-2>(2+4\epsilon)^2-2>2+8\epsilon$$
$$a_4=a_3^2-2>(2+8\epsilon)^2-2>2+16\epsilon$$
$$...$$
$$a_n>2+2^n\epsilon$$
As $n\rightarrow\infty$, $2^n\epsilon\rightarrow\infty$, so $a_n$ clearly diverges.
