Coordinate free proof for $a\times (b\times c) = b(a\cdot c) - c(a\cdot b)$ The vector triple product is defined as $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$. This is often re-written in the following way:
\begin{align*}\mathbf{a}\times (\mathbf{b}\times \mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})\end{align*}
This is a very useful identity for integrating over vector fields and so on (usually in physics).
Every proof I have encountered splits the vectors into components first. This is understandable, because the cross product is purely a three dimensional construct. However, I'm curious as to whether or not there is a coordinate free proof of this identity. Although I don't know much differential geometry, I feel that tensors and so on may form a suitable framework for a coordinate free proof.

Adapted from my previous proof of $\nabla \times (\vec{A} \times \vec{B})$:
\begin{align}
     \vec a \times (\vec b \times \vec c)
 & = a_l \hat{e}_l \times (b_i c_j \hat{e}_k \epsilon_{ijk}) \\
 & = a_l b_i c_j \epsilon_{ijk} \underbrace{ (\hat{e}_l \times \hat{e}_k)}_{(\hat{e}_l \times \hat{e}_k) = \hat{e}_m \epsilon_{lkm} } \\
 & = a_l b_i c_j \hat{e}_m \underbrace{\epsilon_{ijk} \epsilon_{mlk}}_{\text{contracted epsilon identity}} \\
 & = a_l b_i c_j \hat{e}_m \underbrace{(\delta_{im} \delta_{jl} - \delta_{il} \delta_{jm})}_{\text{They sift other subscripts}} \\
 & = a_j (b_i c_j \hat{e}_i) - a_i (b_i c_j \hat{e}_j) \\
 & = (b_i \hat{e}_i) (a_j c_j) - (c_j \hat{e}_j) (a_i b_i) \\
 & = \vec b (\vec a\cdot\vec c) - \vec c(\vec a\cdot \vec b)
\end{align}
