Closed form for an infinite product The following fascinating formula appears in the paper "On gamma quotients and infinite products" by M.Chamberland and A.Straub (see page 9):
$$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$
Unfortunately, it's given without any reference or explanation. Could anyone give one? 

This formula is an application of the theorem $1.1$ for $n=2$ from your Chamberland & Straub paper (when $a+b=c+d$) :
$$\tag{1}\prod_{k\ge 0}\frac {(k+a)(k+b)}{(k+c)(k+d)}=\frac {\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}$$
Observe first that : $\quad\displaystyle n^2-n-1=\left(n-\frac{\sqrt{5}+1}2\right)\left(n+\frac{\sqrt{5}-1}2\right)\;$
An idea is to use $\quad a:=-\dfrac{\sqrt{5}+1}2,\;b:=\dfrac{\sqrt{5}-1}2$.
note that $\;a+b=-1\;$ and take $c=0,\;d=-1$ but we would have $\;0\cdot (-1)\;$ at the denominator so let's shift all these values with an offset $2$ (i.e. set $n:=k+2$) then :
$$a:=2-\frac{\sqrt{5}+1}2,\;b:=2+\frac{\sqrt{5}-1}2,\;c:=2,\;d:=1$$
we then get :
\begin{align}
\prod_{k\ge 0}\frac {(k+2)^2-(k+2)-1}{(k+2)(k+1)}&=\frac {\Gamma(2)\Gamma(1)}{\Gamma\left(2-\frac{\sqrt{5}+1}2\right)\Gamma\left(2+\frac{\sqrt{5}-1}2\right)}\\
&=\frac 1{\left(1-\frac{\sqrt{5}+1}2\right)\left(1+\frac{\sqrt{5}-1}2\right)\Gamma\left(1-\frac{\sqrt{5}+1}2\right)\Gamma\left(1+\frac{\sqrt{5}-1}2\right)}\\
\tag{2}&=-\frac 1{\Gamma\left(1-\frac{\sqrt{5}+1}2\right)\Gamma\left(\frac{\sqrt{5}+1}2\right)}
\end{align}
Using the Euler reflexion formula $\displaystyle \Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin(\pi z)}$ for $z=\dfrac {\sqrt{5}+1}2$ allows to get :
$$\prod_{n\ge 1}\frac {(n+1)^2-(n+1)-1}{n(n+1)}=-\frac {\sin\left(\pi(\sqrt{5}+1)/2\right)}{\pi}$$
and conclude that indeed :
$$\tag{3}\prod_{n\ge 1}\left(1-\frac 1{n(n+1)}\right)=-\frac {\sin\left(\pi(\sqrt{5}+1)/2\right)}{\pi}=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right)$$
