I want to prove, given $\textbf{a}\perp\textbf{b}$, $$\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\mid \textbf{a}\mid^{2} \textbf{b}$$
I realize this is just a matter of 
$$\textbf{a}\cdot(\textbf{a}\wedge\textbf{b})=\langle\textbf{aab}\rangle\_{2-1}=\langle\mid\textbf{a}\mid^2\textbf{b}\rangle\_1=\mid\textbf{a}\mid^2\textbf{b}$$
But I would like to know if there is anything wrong with the following line of reasoning:
$$\mid\textbf{a}\mid^2\textbf{b}=(\textbf{aa})\textbf{b}\stackrel{1}{=}\textbf{aab}\stackrel{2}{=}\textbf{a}\cdot \textbf{ab}\stackrel{3}{=}\textbf{a}\cdot (\textbf{ab})\stackrel{4}{=}\textbf{a}\cdot (\textbf{a}\wedge\textbf{b})$$

1: Geometric Product is associative

2: In this case $\textbf{a}\cdot \textbf{a}$ *is* the geometric product **aa**

3: Geometric product takes precedent in the order of operations (Macdonald LAGA pg. 82)

4: In this case $\textbf{a}\wedge \textbf{b}$ *is* the geometric product **ab**