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? 

We will be only dealing with the following identities:
$$
\Gamma (z) = \frac{ e^{ -\gamma z }}{ z } \, \prod_{ k=1 }^{ \infty  } \left\{ \left( 1+\frac{ z }{ k }  \right)^{ -1 } \, e^{ \frac { z }{ k }   } \right\}  
\hspace{10mm}
\Gamma (-z)\Gamma (z)\quad =\quad \frac { -\pi  }{ z \sin { (\pi z) }  } 
$$
Let's start then ^^:
\begin{align}
\prod_{ n=2 }^{ \infty  }{ \left( 1-\frac { 1 }{ n(n-1) }  \right)  } &=  \prod_{ n=2 }^{ \infty  }{ \left( \frac { { n }^{ 2 }-n-1 }{ n(n-1) }  \right)  } \\ 
&= \prod_{ n=2 }^{ \infty  }{ \left( \frac { (n-{ \phi  }_{ + }) }{ n }  \right)  } \prod _{ n=2 }^{ \infty  }{ \left( \frac { (n-{ \phi  }_{ - }) }{ (n-1) }  \right)  } \\ 
&= \prod_{ n=2 }^{ \infty  }{ \left( 1-\frac { { \phi  }_{ + } }{ n }  \right)  } \prod _{ k=1 }^{ \infty  }{ \left( \frac { k+1-{ \phi  }_{ - } }{ k }  \right)  } \\
&= \frac { \prod _{ k=1 }^{ \infty  }{ \left( 1-\frac { { \emptyset  }_{ + } }{ k }  \right)  }  }{ 1-{ \phi  }_{ + } } \prod _{ k=1 }^{ \infty  }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k }  \right)  } 
\end{align}
Now let's calculate each product using our identity:
\begin{align}
\prod_{ k=1 }^{ \infty  }{ \left( 1+\frac { 1-{ \phi  }_{ - } }{ k }  \right) { e }^{ -\frac { 1-{ \phi  }_{ - } }{ k }  } } 
&= \frac { { e }^{ -\gamma (1-{ \phi  }_{ - }) } }{ (1-{ \phi  }_{ - })\Gamma (1-{ \phi  }_{ - }) } \\ 
\prod_{ k=1 }^{ \infty  }{ \left( 1-\frac { { \phi  }_{ + } }{ k }  \right)  } { e }^{ \frac { { \phi  }_{ + } }{ k }  } 
&= \frac { { e }^{ \gamma { \phi  }_{ + } } }{ -{ \phi  }_{ + }\Gamma (-{ \phi  }_{ + }) } 
\end{align}
Now we will multiply both the products(observe that ${ \phi  }_{ + }+{ \phi  }_{ - } = 1$):
\begin{align}
\prod_{ k=1 }^{ \infty  }{ \left( 1+\frac { 1-{ \phi  }_{ - } }{ k }  \right)  } { e }^{ \frac { { \phi  }_{ - }-1 }{ k }  }\prod _{ k=1 }^{ \infty  }{ \left( 1-\frac { { \phi  }_{ + } }{ k }  \right)  } { e }^{ \frac { { \phi  }_{ + } }{ k }  }
&= \prod _{ k=1 }^{ \infty  }{ \left( 1+\frac { 1-{ \phi  }_{ - } }{ k }  \right) \left( 1-\frac { {\phi  }_{ + } }{ k }  \right)  } \\ 
&= \frac { 1 }{ -\Gamma (-{ \phi  }_{ + })\Gamma (1-{ \phi  }_{ - }){ \phi  }_{ + }(1-{ \phi  }_{ - }) } 
\end{align}
Observe that:
$$
\Gamma (-{ \phi  }_{ + })\Gamma (1-{ \phi  }_{ - }) = \Gamma (-{ \phi  }_{ + })\Gamma ({ \phi  }_{ + })  = \frac { -\pi \csc { (\pi { \phi  }_{ + }) }  }{ { \phi  }_{ + } } 
$$
so:
$$
-\Gamma (-{ \phi  }_{ + })\Gamma (1-{ \phi  }_{ - }){ { \left( { \phi  }_{ + } \right)  }^{ 2 } } = \pi \csc { (\pi { \phi  }_{ + }){ \phi  }_{ + } } 
$$
We get:
\begin{align}
\prod_{ n=1 }^{ \infty  }{ \left( 1-\frac { 1 }{ n(n-1) }  \right)  } 
&= \frac { \prod _{ k=1 }^{ \infty  }{ \left( 1-\frac { { \phi  }_{ + } }{ k }  \right)  } \prod _{ k=1 }^{ \infty  }{ \left( 1+\frac { 1-{ \phi }_{ - } }{ k }  \right)  }  }{ 1-{ \phi  }_{ + } } \\ 
&= \frac { \sin { (\pi { \phi }_{ + }) }  }{ \pi { \phi  }_{ + }(1-{ \phi }_{ + }) } \\
&= -\frac { \sin { (\frac { \pi  }{ 2 } +\frac { \sqrt { 5 }  }{ 2 } \pi ) }  }{ \pi  } \\  
&= -\frac { \cos { \left( \frac { \sqrt { 5 } \pi  }{ 2 }  \right)  }  }{ \pi  } 
\end{align}
