I would like to show that
$$ \begin{align} F_{\alpha i} F_{\beta j} \cdots F_{\gamma k} e_\alpha \wedge e_\beta \cdots \wedge e_\gamma I^\dagger &= e_{\alpha} \cdot F(e_i) e_\beta \cdot F(e_j) \cdots e_\gamma \cdot F(e_k) e_\alpha \wedge e_\beta \cdots \wedge e_\gamma I^\dagger \\ &= F(e_i) \wedge F(e_j) \cdots F(e_k) I^\dagger \end{align} $$
where $F(e_i)$ is a linear function of the vector $e_i$, and $F_{ij} \equiv e_i \cdot F(e_j)$. The $\{e_k\}$ are all orthogonal frame vectors. This is stated in Doran and Lasenby (4.202) pg 117, Geometric Algebra for Physicists.
Attempt:
$$ \begin{align*} e_{\alpha} \cdot F(e_i) e_\beta \cdot F(e_j) \cdots e_\gamma \cdot F(e_k) e_\alpha \wedge e_\beta \cdots \wedge e_\gamma I^\dagger &=(?) \langle e_{\alpha} F(e_i) e_\beta F(e_j) \cdots e_\gamma F(e_k) e_\alpha \cdots e_\gamma I^\dagger \rangle \end{align*} $$
It is not clear how to simplify this. I simply changed the dot products to geometric products under the scalar projection operator, but I'm not sure if this is correct.
Note that Doran and Lasenby say orthonormal frame vectors, not just orthogonal. Since $e_\alpha$ is an orthonormal basis and (from what I can tell) we are working with a Euclidean metric, it is its own reciprocal $e^\alpha$. Thus $$ (e_\alpha\cdot F(e_i))e_\alpha = e_\alpha(e^\alpha\cdot F(e_i)) = F(e_i). $$ This follows simply by definition of the reciprocal basis and by expressing $F(e_i)$ in the basis $e_\alpha$.