On page 40 of Geometric Algebra for Physicists, Doran states in equation 2.98 that for a unit vector $n$ and an arbitrary vector $a$ with $a_\perp = n n\wedge a$, that
$$n\cdot a_\perp=\langle n n \text{ } n\wedge a\rangle=\langle n \wedge a\rangle = 0.$$
Here $\langle \rangle$ is the $0$-grade part of a multi vector (scalar part). He claims this is a proof that $a_\perp$ is perpendicular to $n$.
Yet, on page 38 he recognizes that inner and outer products are always performed before geometric products. Here, he seems to be defying that convention by using $n^2=1$ and performing that geometric product before the wedge product. Is this an error?
This is technically an error because he introduces a new convention of separating two products he wants to perform left-to-right with a space, but never explicitly states the convention. What he really means is
$$n\cdot a_\perp=\langle (n n) n\wedge a\rangle=\langle n \wedge a\rangle = 0,$$
Or, to be more precise by not skipping steps:
$$n\cdot a_\perp=\langle n(n(n\wedge a))\rangle=\langle (n n) \text{ } n\wedge a\rangle=\langle n \wedge a\rangle = 0,$$
With the second equality following from the associative law.