Vector Times Magnitude Same Length As Magnitude Times Vector

Theorem
Given two vectors $\vec{u}$ and $\vec{v}$ of length $\left\|{\vec{u}}\right\|$ and $\left\|{\vec{v}}\right\|$ respectively, $\left\|{\left({\vec{u}\left\|{\vec{v}}\right\|}\right)}\right\| = \left\|{\left({\left\|{\vec{u}}\right\|\vec{v}}\right)}\right\|$.

That is, the first vector times the length of the second vector has the same length as the length of the first vector times the second vector.

Proof
Let $\vec{u} = \left({u_1,u_2,\ldots,u_n}\right)$ and $\vec{v} = \left({v_1,v_2,\ldots,v_n}\right)$.

Note that $\vec{u}\left\|{\vec{v}}\right\| = \left({u_1 \left\|{\vec{v}}\right\|,u_2 \left\|{\vec{v}}\right\|,\ldots,u_n \left\|{\vec{v}}\right\|}\right)$.