# Outer product for r vectors in terms of the geometric product

I found a definition in (An introduction to geometric algebra and calculus by Alan Bromborsky) which say
let $$a_1,a_2,...,a_r$$ be vectors in $$\mathbb{R}^n$$ and let $$\varepsilon_{1...r}^{i_1...i_r}$$ be mixed permutation symbol then$$a_1\wedge ...\wedge a_r =\frac{1}{r!}\sum_{i_1,...,i_r}\varepsilon_{1...r}^{i_1...i_r}a_{i_1}...a_{i_r}$$ And let $$e_1,e_2,...,e_r$$ be an orthogonal basis for the set of linearly independent vectors $$a_1,a_2,...,a_r$$ so that we can write $$a_i=\sum_{j}\alpha_{ij}e_j$$ then geometric product define as $$a_1a_2...a_r=(\sum_{j_1}\alpha_{1j_1}e_{j_1})(\sum_{j_2}\alpha_{2j_2}e_{j_2})...(\sum_{j_r}\alpha_{rj_r}e_{j_r})$$ $$=\sum_{j_1,j_2,...,j_r}\alpha_{1j_1}\alpha_{2j_2}...\alpha_{rj_r}e_{j_1}e_{j_2}...e_{j_r}$$
our aim is to put second definition in the first one.
my problem in this defintion how we can be sure that our outer product has a grade r or in other word the repeated factors of orthogonal basis will vanish

Orthogonal basis elements anticommute, so when you take an antisymmetrized sum over them duplicates vanish. Antisymmetrization is equivalent to alternation: if $$f(x_1,\dotsc,x_k )$$ is any multilinear function then $$f\text{ is antisymmetric} \iff f(x_1,\dotsc,x_k) = 0\text{ whenever }x_i = x_j\text{ for some }i\ne j.$$ Work this out in the case that $$k = 2$$: for $${\implies}$$ consider that $$f(x,x) = -f(x,x)$$; for $${\impliedby}$$ consider that $$f(x_1 + x_2, x_1 + x_2) = 0$$. This idea extends easily to arbitrary $$k$$.
More concretely, in $$X = (a_1e_1 + a_2e_2)(b_1e_1 + b_2e_2) - (b_1e_1 + b_2e_2)(a_1e_1 + a_2e_2)$$ the diagonal terms vanish because e.g. $$a_1b_1e_1e_1 - b_1a_1e_1e_1 = 0.$$ The cross terms combine by anticommuting basis elements: $$a_1b_2e_1e_2 - b_2a_1e_2e_1 = a_1b_2e_1e_2 + b_2a_1e_1e_2 = (a_1b_2 + b_2a_1)e_1e_2.$$ All together you will find $$X = (a_1b_2 - a_2b_1 - [b_1a_2 - b_2a_1])e_1e_2 = 2(a_1b_2 - a_2b_1)e_1e_2.$$