It's known that  

$$S\_2={2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$ 


The terms in the product approaches 1, the same holds for the following convergent series, with $\phi$ the golden ratio 


$$S\_1 = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$ 


Let $$S\_n = {\sqrt{n}\over c\_n}{\sqrt{n+\sqrt{n}}\over c\_n}\dots$$ 


Where $c\_n$ is the solution to the equation $x=\sqrt{n+x}$ 


Is there a simpler formula for $S\_n$? 


What is the asymptotic behavior (Big-O) of $S\_n$ as $n->\infty$? 



Generalized Vieta-product
I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine.


Let $n$ be fixed and $c=c\_n$. Notice that $n=c^2-c$ and $c\to\infty$ as soon as $n\to\infty$.


Denote by $p\_k$ the $k$-th multiplier in the product $S\_n$. It can be easily seen that
$$c\cdot (p\_k^2-1) = p\_{k-1} - 1$$


Consider the functional equation:
$$c\cdot(f(x)^2-1)=f(2cx)-1$$ 
with $f(0)=1$ and $f'(0)=1$. Its solution can be expressed as a series:
$$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots.$$
Then
$$p\_k = f\left(\frac{x\_0}{(2c)^k}\right)$$
where $x\_0$ is a solution to $f(x\_0)=0$.


Now 
$$S\_n = \prod\_{k=1}^{\infty} p\_k = \exp \sum\_{k=1}^{\infty} \ln\left(1 + \Theta\left(\frac{x\_0}{(2c)^k}\right) \right) = \exp \sum\_{k=1}^{\infty} \Theta\left(\frac{x\_0}{(2c)^k}\right) = \exp \Theta\left(\frac{x\_0}{2c-1}\right)$$
which tends to $1$ as $c\to\infty$.


Therefore, $S\_n\to 1$ as $n\to\infty$.


**Example**. For $n=2$, the functional equation admits the analytic solution $f(x)=\cosh(\sqrt{2x})$ for which $x\_0=\frac{-\pi^2}{8}$.


