Answer
Using Eq. (1) and Eq. (3) of Theorem 1 of Section 14.4 we prove that
${a_{\bf{N}}} = \frac{{||{\bf{a}} \times {\bf{v}}||}}{{||{\bf{v}}||}}$
Work Step by Step
By Eq. (1) we have ${a_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2}$.
By Eq. (3) of Theorem 1 of Section 14.4, the curvature is given by
$\kappa \left( t \right) = \frac{{||{\bf{r}}'\left( t \right) \times {\bf{r}}{\rm{''}}\left( t \right)||}}{{||{\bf{r}}'\left( t \right)|{|^3}}}$
or
$\kappa \left( t \right) = \frac{{||{\bf{v}}\left( t \right) \times {\bf{a}}\left( t \right)||}}{{||{\bf{v}}\left( t \right)|{|^3}}} = \frac{{||{\bf{a}}\left( t \right) \times {\bf{v}}\left( t \right)||}}{{||{\bf{v}}\left( t \right)|{|^3}}}$
Since $||{\bf{v}}\left( t \right)|| = v\left( t \right)$, thus,
${a_{\bf{N}}} = \frac{{||{\bf{a}}\left( t \right) \times {\bf{v}}\left( t \right)||}}{{||{\bf{v}}\left( t \right)|{|^3}}}v{\left( t \right)^2} = \frac{{||{\bf{a}}\left( t \right) \times {\bf{v}}\left( t \right)||}}{{||{\bf{v}}\left( t \right)||}}$
We drop the parameter $t$ for convenience and obtain ${a_{\bf{N}}} = \frac{{||{\bf{a}} \times {\bf{v}}||}}{{||{\bf{v}}||}}$
