I'm trying to prove that $$(\vec a \times \vec b) \times (\vec a \times \vec c)=\vec a(\vec a \cdot \vec b \times \vec c)$$ using index notation (i.e. Einstein sumnmation notation).
Here's what I've done so far: $$\begin{align} (\vec a \times \vec b) \times (\vec a \times \vec c) &=\varepsilon_{ijk}a_jb_k\varepsilon_{ilm}a_lc_m \\ \ \\&=[\varepsilon_{ijk}\varepsilon_{ilm}]a_ja_lb_kc_m \\ \ \\&=[\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}]a_ja_lb_kc_m \\ \ \\&=(\delta_{jl}a_j)(\delta_{km}b_k)a_lc_m-(\delta_{jm}a_j)(\delta_{kl}b_k)a_lc_m \\ \ \\&=a_lb_ma_lc_m-a_mb_lc_m \\ \ \\&=a_l[a_lb_mc_m-a_mb_lc_m] \\ \ \\&=\vec a[(\vec b \cdot \vec c)\vec a-(\vec a \cdot \vec c) \vec b],\end{align}$$ but I'm stuck on how to proceed from here.
Any hints/tips?
P.S. I'm trying to do this just using index notation, so I want to avoid vector identities (like the one for $\vec a \cdot \vec b \times \vec c$).
Your first equation is wrong: You left out the Levi-Civitta symbol coming from the middle cross product (and thus end up with an expression with no free indices, which you know can't be right).
The correct starting point is
$$ \epsilon_{krp} (\epsilon_{ijk}a_ib_j)\epsilon_{mnr}a_mc_n) $$ which simplifies as follows: $$ \begin{array}{c} \epsilon_{mnr} (\epsilon_{krp} \epsilon_{ijk})a_ib_ja_mc_n =\\ \epsilon_{mnr} (\delta_{ri}\delta_{pj}-\delta_{rj}\delta_{pi})a_ib_ja_mc_n =\\ \epsilon_{mnr} \left( a_ra_mb_pc_n - a_pa_mb_rc_n\right) =\\ -\epsilon_{mnr} a_pa_mb_rc_n = (\epsilon_{mrn} a_mb_rc_n) a_p = \\ \left( a_m [\epsilon_{mrn} b_rc_n] \right) a_p =\\ \left( \vec{a}\cdot (\vec{b}\times\vec{c} \right) \vec{a} \end{array} $$