The magnitude of a triple product of two vectors So I was going through a past exam for electrodynamics and a question for radiation came up and within it was the following magnitude of a triple product
$ \lvert \hat{r} \times [\hat{r} \times \vec{\beta} ] \rvert^2 $
I looked in the solutions later to check I was correct and they gave the following for how to do that nasty magnitude: 
$ \lvert \hat{r} \times [\hat{r} \times \vec{\beta} ] \rvert^2 = \lvert \hat{r} \times \vec{\beta} \rvert^2$ 
I can't understand how they obtained this formula I've tried going through the triple product BAC - CAB identity but I just don't see it and it seems super useful to have. I was wondering if someone could give me how to obtain this.

First start with the double vector product identity :
$$A\times(A\times B) = (A \cdot B) A - (A \cdot A) B$$
Then, you get, using the cyclicity of the mixt product :
$$ \begin{array}{rcl}
\|A\times B\|^2 & = & (A\times B) \cdot (A \times B) \\
 & = & B \cdot ((A\times B) \times A) \\
 & = & B \cdot \left[ (A \cdot A) B - (A \cdot B) A \right] \\
 & = & A^2 B^2 - (A \cdot B)^2
\end{array}$$
Then, you have :
$$\begin{array}{rcl}
\|A\times(A\times B)\|^2 & = & \left[ (A \cdot B) A - (A \cdot A) B \right]^2 \\
 & = & (A\cdot B)^2 A^2 + A^4 B^2 -2 A^2(A \cdot B)^2 \\
 & = & A^2 \left[ A^2 B^2 - (A \cdot B)^2 \right] \\
 & = & A^2 \|A \times B\|^2
\end{array}$$
And, in your case, you conclude with $r$ being unitary.
