= Solution
Using <Einstein summation convention> and the <contraction of two Levi-Civita symbols>,
$$
\begin{aligned}
[a\times(b\times c)]_i
&=\epsilon_{ijk}\epsilon_{k\ell m}a_jb_\ell c_m\\
&=(\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})a_jb_\ell c_m\\
&=b_i(a\cdot c)-c_i(a\cdot b).
\end{aligned}
$$
This proves the <vector triple product identity> directly from the <Levi-Civita symbol> and <Kronecker delta>.
Define the <scalar triple product> by $[a,b,c]=a\cdot(b\times c)$. Its expression $\epsilon_{ijk}a_ib_jc_k$ is unchanged by cyclically relabeling the three indices: a three-cycle has positive sign. Thus
$$
[a,b,c]=[b,c,a]=[c,a,b].
$$
In particular, let $\Delta=[e_1,e_2,e_3]$. Since the three vectors form a <basis>, their <determinant> $\Delta$ is nonzero, so their <reciprocal basis> vectors are well defined. Cyclic invariance and vanishing of a <scalar triple product> with repeated vectors give
$$
\widehat e_i\cdot e_j=\delta_{ij}.
$$
To calculate the new <scalar triple product> explicitly, apply the already proved <vector triple product identity> with $A=e_3\times e_1$:
$$
\begin{aligned}
\widehat e_2\times\widehat e_3
&=\frac{(e_3\times e_1)\times(e_1\times e_2)}{\Delta^2}\\
&=\frac{[(e_3\times e_1)\cdot e_2]e_1-[(e_3\times e_1)\cdot e_1]e_2}{\Delta^2}
=\frac{e_1}{\Delta}.
\end{aligned}
$$
Consequently
$$
\boxed{[\widehat e_1,\widehat e_2,\widehat e_3]=\frac1\Delta\neq0,}
$$
so the reciprocal vectors also form a <basis>. Taking the <reciprocal basis> of that <basis> now gives
$$
\widehat{\widehat e}_1
=\frac{\widehat e_2\times\widehat e_3}{[\widehat e_1,\widehat e_2,\widehat e_3]}
=\frac{e_1/\Delta}{1/\Delta}=e_1.
$$
The same cyclic calculation gives $\widehat{\widehat e}_2=e_2$ and $\widehat{\widehat e}_3=e_3$, proving <reciprocal basis involution>.
Finally, write $V=V_1e_1+V_2e_2+V_3e_3$. Taking a <dot product> with $\widehat e_j$ isolates $V_j$. Applying the same argument to the <reciprocal basis>, whose reciprocal is the original <basis>, gives the two sets of coordinates:
$$
\boxed{V_j=V\cdot\widehat e_j,\qquad \widehat V_j=V\cdot e_j.}
$$
No <orthonormality> assumption is needed; the <reciprocal basis> is what makes these coordinate-extraction formulas work.
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