A high-level reason that $u \cdot (v \times w) = (u \times v) \cdot w$? I can do the algebra to show that for $u, v, w \in \mathbb{R}^3$, this identity is true:
$$u \cdot (v \times w) = (u \times v) \cdot w$$
But is there a more high-level reason? I didn't expect the cross and dot product to be connected in this surprising way.
 A: $$u \cdot (v \times w) = \det(u, v, w) $$
where (u, v, w) means the 3x3 matrix whose columns are $u$, $v$, and $w$ respectively, and thus the triple product satisfies all of the algebraic identities that determinants do.
(alternatively, $(u,v,w)$ can mean the matrix whose rows are $u$, $v$, and $w$, if you prefer to think of vectors as row vectors)
A: Yes. The volume of the parallelepipid is equal to the scalar triple product. Consider the geometric definitions of the cross-product and inner-product to see this.
A: $u \times v$ can be defined as the vector that makes 
$$\langle w, u \times v \rangle=\det(w,u,v) \quad  \forall w.$$
That there is such a vector can be seen by evaluating the above equality with the canonical basis. That it is unique can be seen in the same way. Now your result follows easily.
A: $u\cdot(v\times w)$ is the determinant of the matrix whose columns are $u$, $v$ and $w$. (This can be used as a definition for the cross product).
The identity now follows from the usual determinant rules for permuting rows/columns.
