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](http://geocalc.clas.asu.edu/pdf/Tutorial%20on%20Geometric%20Calculus.pdf) by David Hestenes. Can anyone help me to derive it? I thought that: $\nabla \cdot(\mathbf{a} \wedge\mathbf{r}) = \epsilon^{ijk}\partial\_i a\_j x\_k = 0$.