Answer
${\bf{e}} \times \left( {{\bf{e}}' \times {\bf{e}}} \right) = {\bf{e}}'$
Work Step by Step
Using the geometric description of the cross product given by Theorem 1, we have
(i) ${\bf{e}} \bot {\bf{e}}'$ is orthogonal to ${\bf{e}}$ and ${\bf{e}}'$.
Let us define the vector ${\bf{e}}{\rm{''}}$ such that ${\bf{e}}{\rm{''}} = {\bf{e}}' \times {\bf{e}}$.
(ii) ${\bf{e}}{\rm{''}}$ has length $||{\bf{e}}||||{\bf{e}}'||\sin \theta $, where $\theta$ is the angle between ${\bf{e}}$ and ${\bf{e}}'$ and $0 \le \theta \le \pi $.
Since ${\bf{e}}$ and ${\bf{e}}'$ are unit vectors and ${\bf{e}} \bot {\bf{e}}'$, so ${\bf{e}}{\rm{''}}$ is also a unit vector.
Thus, ${\bf{e}} \times \left( {{\bf{e}}' \times {\bf{e}}} \right)$ is a unit vector.
(iii) $\left\{ {{\bf{e}}',{\bf{e}},{\bf{e}}{\rm{''}}} \right\}$ forms a right-handed system.
Thus, by the right-hand rule ${\bf{e}} \times {\bf{e}}{\rm{''}} = {\bf{e}} \times \left( {{\bf{e}}' \times {\bf{e}}} \right) = {\bf{e}}'$.
