Wedge product and differential forms

Question

I am a bit confused when it comes to wedge product and differential forms.

I know the following property:

$\omega\wedge\eta=(−1)^{kl}\eta\wedge\omega$

Also I know that when $k$ is odd and I am doing the wedge product of the same vector the answer is zero.

Ex:

$\omega\wedge\eta=(−1)^{kk}\omega\wedge\omega$

$-\omega\wedge\omega=0$

Here is my question when calculating the following:

$(\omega\wedge\eta) \wedge \nu =(3(dx \wedge dy) − 2(dz \wedge dx) − 7(dy \wedge dz)) \wedge (2dx + 3dy + 2dz)$

$=6(dx∧dy∧dx)+9(dx∧dy∧dy)+6(dx∧dy∧dz) −4(dz∧dx∧dx)−6(dz∧dx∧dy)−4(dz∧dx∧dz) −14(dy∧dz∧dx)−21(dy∧dz∧dy)−14(dy∧dz∧dz) $

$=6(dx ∧ dy ∧ dz) − 6(dz ∧ dx ∧ dy) − 14(dy ∧ dz ∧ dx)$

$=6(dx ∧ dy ∧ dz) \underline- 6(dx ∧ dy ∧ dz) \underline- 14(dx ∧ dy ∧ dz)$

$= − 14(dx ∧ dy ∧ dz)$

Shouldn't the $\underline-$ be turned into positives since we are switching the order of $dx$,$dy$ and $dz$.

$=6(dx ∧ dy ∧ dz) \underline+ 6(dx ∧ dy ∧ dz) \underline+ 14(dx ∧ dy ∧ dz)$.

Does such a change only occur in 2-form but not in 3?

Answer 1

We did two switches: for example $dz \wedge dx \wedge dy = -dx \wedge dz \wedge dy = dx \wedge dy \wedge dz$.