Proving det(A) property with outermorphism definition 
Let A be a square matrix. Prove that exchanging two columns of A changes the sign of $\det(A)$. **Note: I'm pretty sure this is supposed to be "adjacent columns"

Source: "Linear and Geometric Algebra" - Alan MacDonald
The definition of $\det(A)$ that I have is $\det(A)=\det(f)$ where $[f]_\epsilon = A = \left[f(\mathbf e_1)\ f(\mathbf e_2)\ ...\ f(\mathbf e_n)\right]$ and $\det(f)=[f(\mathbf e_1)\wedge f(\mathbf e_2)\wedge\ ...\wedge\ f(\mathbf e_n)] / (\mathbf e_1\wedge \mathbf e_2\wedge\ ...\wedge\ \mathbf e_n)$.
Exchanging adjacent columns in A then will exchange $f(\mathbf e_i)$ with $f(\mathbf e_{i+1})$ in the above outer product.  But even though, $\{\mathbf e_i\}$ is an orthogonal set, $\{f(\mathbf e_i)\}$ isn't necessarily (they are linearly independent, though).  So I can't just say that they're anticommutative and will negate $\det(A)$ upon exchanging adjacent $f(\mathbf e_i)$'s.  So I don't know how to prove this.
 A: Wedge products always anticommute; you do not need an orthogonal basis for that to be true.  So swapping adjacent columns will indeed change the sign.
But it is true that any column swap will change the sign.  For instance, if we swap the first and third columns, we get
$$f(e_3) \wedge f(e_2) \wedge f(e_1) \wedge \ldots = (-1)^3 f(e_1) \wedge f(e_2) \wedge f(e_3) \wedge \ldots$$
The number of swaps needed to get back to "canonical" ordering of the basis is always an odd number.  In particular, if the difference in the column numbers is $n$, then the number of swaps is $2n-1$.
A: Beyond all the notation, you're using the definition $$\det(A)=\sum_{\sigma \in S_n}{\rm sgn}\; \sigma\; a_{1\sigma (1)}\cdots a_{n\sigma (n)}$$
Now suppose we swap columns $i,j$. This amounts to applying $\tau=(ij)$ to the above, in the following sense $$\det(A')=\sum_{\sigma \in S_n}{\rm sgn}\; \sigma\; a_{1\tau\sigma (1)}\cdots a_{n\tau\sigma (n)}$$
Now $\sigma \mapsto \tau\sigma$ is a biyection in $S_n$ and $\tau\tau=1$ so   $$\det(A')=\sum_{\sigma \in S_n}{\rm sgn}\; \tau(\tau\sigma)\; a_{1\tau\sigma (1)}\cdots a_{n\sigma (n)} \\= \sum_{\sigma' \in S_n}{\rm sgn}\; \tau \sigma'\; a_{1\sigma'(1)}\cdots a_{n\sigma'(n)}$$
Now ${\rm sgn}\;\tau\sigma'={\rm sgn}\;\tau\cdot {\rm sgn}\;\sigma'=-{\rm sgn}\;\sigma'$ so that  $$\det(A')=  -\sum_{\sigma' \in S_n}{\rm sgn}\;\sigma'\; a_{1\sigma'(1)}\cdots a_{n\sigma'(n)}=-\det(A)$$
