derivative formula $\nabla \times (\mathbf{a} \times \mathbf{r}) = \nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = (n-1)\mathbf{a}$ Assume $\mathbf{r}=\mathbf{x}−\mathbf{x}′$ is the position vector in $\mathbb{R}^n$, for constant $\mathbf{a}$, we have  
$$\nabla \times (\mathbf{a} \times \mathbf{r}) = \nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = (n-1)\mathbf{a}.$$ 
This comes from fig.6 in Tutorial on Geometric Calculus by David Hestenes. Can anyone help me to derive it? I thought that: $\nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = \epsilon^{ijk}\partial_i a_j x_k = 0$.

I would think it is still applicable for traditional vector analysis approach. With $\mathbf{r}\in\mathbb{R}^n$, then
\begin{align}
[\nabla\times(\mathbf{a}\times\mathbf{r})]&=\varepsilon_{ijk}\partial_j[\mathbf{a}\times\mathbf{r}]_k=\varepsilon_{ijk}\partial_j(\varepsilon_{klm}a_{\ell}r_m)\\
&=\varepsilon_{ijk}\varepsilon_{klm}(r_m\partial_ja_{\ell}+a_{\ell}\partial_jr_m)\\
&=(\delta_i^{\ell}\delta_j^m-\delta_i^m\delta_j^{\ell})(r_m\partial_ja_{\ell}+a_{\ell}\partial_jr_m)\\
&=r_j\partial_ja+a\partial_jr_j-r\partial_ja_j-a_j\partial_j{r}\\&=(\mathbf{r}\cdot\nabla)\mathbf{a}+\mathbf{a}(\nabla\cdot\mathbf{r})-\mathbf{r}(\nabla\cdot\mathbf{a})-(\mathbf{a}\cdot\nabla)\mathbf{r}\\
&=\mathbf{a}(\nabla\cdot\mathbf{r})-(\mathbf{a}\cdot\nabla)\mathbf{r}\\
&=n\mathbf{a}-\mathbf{a}\\
&=(n-1)\mathbf{a}
\end{align}
(where I have mixed lower and upper indices).
