Evaluate $\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$ Evaluate $$P=\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$$
My try:
Let $a_k=45^{2^k}$ Then we have $45^{-2^k}=\frac{1}{a_k}$ 
So We get:
$$1+\frac{2}{a_k+\frac{1}{a_k}}=\frac{(a_k+1)^2}{a_k^2+1}$$
So
$$P=\lim_{n \to \infty}\prod_{k=0}^n\frac{(a_k+1)^2}{a_k^2+1}$$
Any way from here

Observe that
Hint:
$$(a_k+1)(a_k-1)=(45^{2^k})^2-1=a_{k+1}-1$$
$$(a_0+1)\prod_{k=1}^n(a_k+1)=a_{n+1}-1$$
Similarly, $$(a_0^2+1)\prod_{k=1}^n(a_k^2+1)=a_{n+1}^2-1$$
