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