I was having trouble simplifying this relation. This is the question.

If $a_0=\frac{5}{2}$ and $a_k=(a_{k-1})^2-2$, find:
 $$\prod_{k=0}^{\infty}\left(1-\frac{1}{a_k}\right).$$

I thought of first finding out some of the elements of this series . My thinking was that the elements would cancel each other in a certain fashion but I was already troubled at $a_2$ which was $\frac{257}{16}$.
Then I thought of simplifying the relation by substituting $a_k$s relation but got stuck . 
So that's where I need help .
Let $ n $ be a positive integer, we have for any $ k\in\mathbb{N} $ : $$ a_{k+1}+1=a_{k}^{2}-1=\left(a_{k}-1\right)\left(a_{k}+1\right)$$

Thus : $$ \prod_{k=0}^{n-1}{\left(a_{k}-1\right)}=\prod_{k=0}^{n-1}{\frac{a_{k+1}+1}{a_{k}+1}}=\frac{a_{n}+1}{a_{0}+1}=\frac{2}{7}\left(a_{n}+1\right) $$

Again, for any $ k\in\mathbb{N} $, we have : $$ a_{k+2}-2=a_{k+1}^{2}-4=\left(a_{k+1}-2\right)\left(a_{k+1}+2\right)=a_{k}^{2}\left(a_{k+1}-2\right) $$

Thus : $$ \prod_{k=0}^{n-1}{a_{k}}=\sqrt{\prod_{k=0}^{n-1}{a_{k}^{2}}}=\sqrt{\prod_{k=0}^{n-1}{\frac{a_{k+2}-2}{a_{k+1}-2}}}=\sqrt{\frac{a_{n+1}-2}{a_{1}-2}}=\frac{2}{3}\sqrt{a_{n+1}-2}=\frac{2}{3}\sqrt{a_{n}^{2}-4} $$

Hence, for every positive integer $ n $, we have : $$ \fbox{$\begin{array}{rcl}\displaystyle\prod_{k=0}^{n-1}{\left(1-\frac{1}{a_{k}}\right)}=\frac{3}{7}\times\frac{a_{n}+1}{\sqrt{a_{n}^{2}-4}}\end{array}$} $$

Let's figure out the limit of $ \left(a_{n}\right) $ :

Let $ n $ be a positive integer.
We can prove by induction that $ \left(\forall k\in\mathbb{N}\right),\ a_{k}\geq 2 $, from which we get that $ \left(\forall k\in\mathbb{N}\right),\ a_{k}^{2}-3\geq\left(a_{k}-1\right)^{2} $, hence $ \left(\forall k\in\mathbb{N}\right),\ a_{k+1}-1\geq\left(a_{k}-1\right)^{2} $, thus $ \left(\forall k\in\mathbb{N}\right),\ \left(a_{k+1}-1\right)^{2^{n-k-1}}\geq\left(a_{k}-1\right)^{2^{n-k}} \cdot $

Telescoping : $$ \prod_{k=0}^{n-1}{\frac{\left(a_{k+1}-1\right)^{2^{n-k-1}}}{\left(a_{k}-1\right)^{2^{n-k}}}}\geq 1\iff a_{n}-1\geq\left(a_{0}-1\right)^{2^{n}} \iff a_{n}\geq\left(\frac{3}{2}\right)^{2^{n}}+1\underset{n\to +\infty}{\longrightarrow}+\infty $$

Thus : $$ \fbox{$\begin{array}{rcl}\displaystyle\lim_{n\to +\infty}{a_{n}}=+\infty\end{array}$} $$
(I still think we can prove that in a much simpler way)

Finally, tending $ n $ to infinity in the closed form that we got for the product, we get : $$ \prod_{n=0}^{+\infty}{\left(1-\frac{1}{a_{n}}\right)}=\frac{3}{7} $$
Referring to the previous answer by @CHAMSI:
$$ \prod_{k=0}^{n-1}{\left(1-\frac{1}{a_{k}}\right)}=\frac{3}{7}\times\frac{a_{n}+1}{\sqrt{a_{n}^{2}-4}} $$

$\Rightarrow \prod_{k=0}^{\infty}\left(1-\frac{1}{a_k}\right)= \lim_{n \rightarrow \infty} \prod_{k=0}^{n-1}{\left(1-\frac{1}{a_{k}}\right)}=\frac{3}{7} \lim_{n \rightarrow \infty} \frac{a_{n}+1}{\sqrt{a_{n}^{2}-4}}.$

As mentioned by @Integrand, we can easily prove by induction that $a_n=2^{2^n}+\frac{1}{2^{2^n}}$, which clearly goes to infinity as $n$ approaches infinity.