Viète's formula expresses an infinite product of nested radicals in terms of $\pi$.
Let $$a_1=\sqrt2,\quad a_n=\sqrt{2+a_{n-1}}.$$ Note that $\lim\limits_{n\to\infty}a_n=2.$ Then $$\prod_{n=1}^\infty\frac{a_n}2=\frac{\sqrt2}2\times\frac{\sqrt{2+\sqrt2}}2\times\frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\times\dots=\frac2\pi.$$ Let's consider a variation of this problem. Let $$b_1=\sqrt1=1,\quad b_n=\sqrt{1+b_{n-1}}.$$ It's easy to see that $\lim\limits_{n\to\infty}b_n=\phi,$ where $\phi=(1+\sqrt5)/2$ is the golden ratio.
Is it possible to find a closed form for this infinite product? $$\begin{align} P&=\prod_{n=1}^\infty\frac{b_n}\phi=\frac{\sqrt1}\phi\times\frac{\sqrt{1+\sqrt1}}\phi\times\frac{\sqrt{1+\sqrt{1+\sqrt1}}}\phi\times\dots\\ &\approx0.5094909728475357551307361756807565841868\dots \end{align}$$ Or, at least, to find alternative representations and establish some of its properties: is it irrational? is it transcendental?