How can $(A \times \nabla) \times B$ be rearranged? Given 3 vector spaces $A$, $B$ & $C$, I am able to derive that $A \times (B \times C)$ is equal to $(A \cdot C)B-(A \cdot B)C$.
However, I do not think I can apply this identity to $(A \times \nabla) \times B$, because I would get:
$$(A \times \nabla) \times B = (A \cdot B)\nabla - (A \cdot \nabla)B$$
Which is clearly wrong, since $(A \cdot \nabla)\nabla$ is not a vector space.
How would I go about this then?

Working with the Levi-Civita symbol yields:
$$\epsilon_{ijk}\epsilon_{jlm}A_l\partial_mB_k=A_k \partial_i B_k - A_i\partial_kB_k $$
Which doesn't seem to be particularly useful in this case.
 A: Here's an approach that I like.  First, note that we can write
$$
\nabla \times F = \sum_{i=1}^3 e_i \times \frac{\partial F}{\partial x_i}
$$
where $e_1,e_2,e_3$ denote $\hat i, \hat j, \hat k$, and 
$$
\frac{\partial F}{\partial x_i} = \left(\frac{\partial F^{(x_1)}}{x_i},\frac{\partial F^{(x_2)}}{x_i}, \frac{\partial F^{(x_3)}}{x_i}\right).
$$ 
With that in mind, we can write
$$
\begin{align*}
(A \times \nabla) \times B &= \sum_{i=1}^3 (A \times e_i) \times \frac{\partial B}{\partial x_i} 
\\ & = 
\sum_{i=1}^3 \left[\left(A \cdot \frac{\partial B}{\partial x_i}\right) e_i - \left (\frac{\partial B}{\partial x_i} \cdot e_i \right)A\right]
\\ & =
\sum_{i=1}^3 \left(A \cdot \frac{\partial B}{\partial x_i}\right) e_i - A\sum_{i=1}^3\frac{\partial B}{\partial x_i} \cdot e_i
\\ & =
\sum_{i=1}^3 \left(A \cdot \frac{\partial B}{\partial x_i}\right) e_i - A(\nabla \cdot B)
\end{align*}
$$
According to the other answer, it seems that the first term can be rewritten as $A \cdot (\nabla B)$.
A: We have (using Einstein summation notation):
\begin{align*}
(\mathbf{A}\times\nabla)_i&=\varepsilon_{ijk}A_j\partial_k \\
[(\mathbf{A}\times\nabla)\times \mathbf{B}]_{m}&=\varepsilon_{mil}\underbrace{\varepsilon_{ijk}A_j\partial_k}_{(\mathbf{A}\times\nabla)_i}B_l \\
&=-\varepsilon_{iml}\varepsilon_{ijk}A_j\partial_kB_l \\
&=-(\delta_{mj}\delta_{lk}-\delta_{mk}\delta_{lj})A_j\partial_kB_l \\
&=(\delta_{mk}\delta_{lj}-\delta_{mj}\delta_{lk})A_j\partial_kB_l \\
&=\delta_{mk}\delta_{lj}A_j\partial_kB_l-\delta_{mj}\delta_{lk}A_j\partial_kB_l \\
&=\delta_{mk}A_j\partial_kB_j-\delta_{mj}A_j\partial_kB_k \\
&=A_j\partial_mB_j-A_m\partial_kB_k,
\end{align*}
as you obtained (though with different dummy variables - an inconsequential difference). The final step is to recognize this in terms of vectors and the various products available to us (dot, scalar, cross). We have
$$(\mathbf{A}\times\nabla)\times \mathbf{B}=\mathbf{A}\cdot(\nabla\mathbf{B})-\mathbf{A}(\nabla\cdot\mathbf{B}). $$
Now, the notation $\nabla\mathbf{B}$ requires some explanation. It's the Jacobian matrix
$$\nabla\mathbf{B}=\left(\frac{\partial B_i}{\partial x_j}\right)_{ij}, $$
a second-rank tensor, so that the dot product $\mathbf{A}\cdot(\nabla\mathbf{B})$ is still a vector, as we need (the result should definitely be a vector!). 
