The wedge product of a $1$-form and a $2$-form I'm trying to understand the wedge product of a one form and a two form. My difficulty is that I end up with a $dx \wedge dx \wedge dy$, for example, and I know a wedge of itself is 0. Does this become $0 \wedge dy$, and doesn't that just equal 0? The problem that I'm told that what I end up with should be a three form.
Here is the example I'm working with:

Find $\alpha \wedge \omega$ given that
  $\alpha = dx +dz$ and $\omega = dy \wedge dz + xdz \wedge dx + ydx \wedge dy$.

 A: $dx\wedge dx \wedge dy$ does equal $0$.  But that doesn't mean that the wedge product of a $1$-form and a $2$-form is always $0$.  Let's look at an example.  Let $\alpha_1 = xy\ dx + z^2\ dz$ and $\alpha_2 = 2\ dx\wedge dy - \cos(x)\ dy\wedge dz$.  Then the wedge product is 
$$\begin{align}\alpha_1 \wedge \alpha_2 &= (xy\ dx + z^2\ dz)\wedge(2\ dx\wedge dy - \cos(x)\ dy\wedge dz) \\ &= (xy\ dx)\wedge[2\ dx\wedge dy - \cos(x)\ dy\wedge dz] + (z^2\ dz)\wedge[2\ dx\wedge dy - \cos(x)\ dy\wedge dz] \\ &= \big(2xy\ dx\wedge dx\wedge dy - xy\cos(x)\ dx\wedge dy\wedge dz\big)+ \big(2z^2\ dz\wedge dx\wedge dy- z^2\cos(x)\ dz\wedge dy\wedge dz\big) \\ &= (0-xy\cos(x)\ dx\wedge dy\wedge dz)+ (2z^2\ dx\wedge dy\wedge dz - 0) \\ &= (2z^2-xy\cos(x))\ dx\wedge dy\wedge dz\end{align}$$
To make more explicit my use of the associativity and anticommutativity of the wedge product above, I'll work these parts out here: $$\begin{array}{crl} (1)\quad &dx \wedge dx \wedge dy &= (dx \wedge dx)\wedge dy \\ &&= 0\wedge dy \\ &&= 0 \\ \hline(2)\quad &dz\wedge dx\wedge dy &= (dz \wedge dx)\wedge dy \\ &&= -(dx\wedge dz)\wedge dy \\ &&= -dx\wedge (dz\wedge dy) \\ &&= dx\wedge (dy\wedge dz) \\ &&= dx\wedge dy\wedge dz \\ \hline(3)\quad &dz\wedge dy\wedge dz &= (dz \wedge dy) \wedge dz \\ &&= -(dy\wedge dz)\wedge dz \\ &&= -dy\wedge (dz\wedge dz) \\ &&= -dy\wedge 0 \\ &&= 0\end{array}$$
