The vector quadruple product - Index notation I am looking to prove the following using index notation, but I am getting stuck on obtaining the RHS of the statement. 
Thus, I wanted to be 100% sure (by asking you guys) that the RHS of the statement below is correct.
That is:
$(\overrightarrow{a}$
$\times$$\overrightarrow{b})$$\times(\overrightarrow{a}$$\times$$\overrightarrow{c})$ $=$ $\overrightarrow{a}$$(\overrightarrow{a}$$\cdot$$\overrightarrow{b}$$\times$$\overrightarrow{c})$.
Thanks for your time!
 A: Theorem:
$$(a\times b) \times (a\times c) = a(a\cdot (b\times c))$$
Proof:
$$\begin{align}[(a\times b) \times (a\times c)]_d &= \varepsilon_{def}(a\times b)_e(a\times c)_f \\ &= \varepsilon_{def}(\varepsilon_{egh}a_gb_h)(\varepsilon_{fij}a_ic_j) \\ &= \varepsilon_{efd}\varepsilon_{egh}\varepsilon_{fij}a_gb_ha_ic_j \\ &= (\delta_{fg}\delta_{dh} - \delta_{fh}\delta_{gd})\varepsilon_{fij}a_gb_ha_ic_j \\ &= \delta_{fg}\delta_{dh}\varepsilon_{fij}a_gb_ha_ic_j - \delta_{fh}\delta_{gd}\varepsilon_{fij}a_gb_ha_ic_j \\ &= \varepsilon_{gij}a_gb_da_ic_j - \varepsilon_{hij}a_db_ha_ic_j \\ &= \varepsilon_{gij}a_gb_da_ic_j + \varepsilon_{ihj}a_db_ha_ic_j \\ &= a_g(\varepsilon_{gij}a_ic_j)b_d + a_i(\varepsilon_{ihj}b_hc_j)a_d \\ &= a_g(a\times c)_gb_d + a_i(b\times c)_ia_d \\ &= [a\cdot (a\times c)]b_d + [a\cdot (b\times c)]a_d\end{align}$$

Lemma:
$$a\cdot (a\times c) = 0$$
Proof:
$$\begin{align} a\cdot (a\times c) &= a_i(\varepsilon_{ijk}a_jc_k) \\ &= -a_i\varepsilon_{jik}a_jc_k \\ &= -a_j\varepsilon_{jik}a_ic_k \tag{commuting terms} \\ &= -a_i\varepsilon_{ijk}a_jc_k \tag{renaming my dummy indices: $i \leftrightarrow j$} \\ &= -a_i(a\times c)_i \\ &= -a\cdot (a\times c) \end{align}$$
The only scalar equal to its negative is zero.  Thus $a\cdot (a\times c) = 0$.$\ \ \ \ \ \square$

Using this lemma we see that the first term in our theorem is zero.
Thus $$\begin{align}[(a\times b) \times (a\times c)]_d &= [a\cdot (b\times c)]a_d \\ &= a_d[a\cdot (b\times c)] \\ &= \left[a\left(a\cdot [b\times c]\right)\right]_d\end{align}$$
Because this equality holds for an arbitrary $d$th coordinate of each vector, it holds for the vectors.$\ \ \ \ \square$
