The projection of vector $\mathbf a$ onto vector $\mathbf v$ is given by $$ \left( \frac {\mathbf a \cdot \mathbf v} {\left\lVert \mathbf v^2 \right\rVert} \right) \mathbf v$$
This seems like a roundabout way to solve the problem. Can't the projection of vector $\mathbf a$ onto vector $\mathbf v$ simply be given by $$ \left( \frac {\mathbf a \cdot \mathbf v} {\left\lVert \mathbf v \right\rVert} \right)$$ Since,
$$ \left( \frac {\mathbf a \cdot \mathbf v} {\left\lVert \mathbf v \right\rVert} \right) = \left( \frac {\left\lVert \mathbf a \right\rVert\left\lVert \mathbf v \right\rVert \cos \theta} {\left\lVert \mathbf v \right\rVert} \right) = \left\lVert \mathbf a \right\rVert \cos \theta$$ where $\theta$ is the angle between vector $\mathbf a$ and vector $\mathbf v$?
If so, can it be claimed that $$ \left( \frac {\mathbf a \cdot \mathbf v} {\left\lVert \mathbf v^2 \right\rVert} \right) \mathbf v = \left( \frac {\mathbf a \cdot \mathbf v} {\left\lVert \mathbf v \right\rVert} \right)$$ And why does all the documentation go by the first rather than the second proof?
The projection, also known as the "resolved part", of vector $\underline{a}$ onto vector $\underline{v}$ is the component of $\underline{a}$ in the direction of $\underline{v}$ and is therefore a scalar quantity, not a vector quantity.
It is correctly given by the formula you gave: $$\frac{\underline{a}\cdot\underline{v}}{|\underline{v}|}$$