On the wedge product of forms Pages 36-37 of Loring Tu's Introduction to Manifolds says:

The wedge product of a $k$-form $\omega$ and an $l$-form $\tau$ on an open set $U$ is defined pointwise:
  $$(\omega\wedge \tau)_p=\omega_p\wedge \tau_p, \hspace{.75cm} p\in U. $$
  In terms of coordinates, if $\omega=\sum_I a_Idx^I$ and $\tau=\sum_J b_J dx^J$, then 
  $$\omega \wedge \tau=\sum_{I,J}(a_Ib_J)dx^I\wedge dx^J\tag{$\ast$}.$$ 
  In this sum, if $I$ and $J$ are not disjoint on the right hand side, then $dx^I\wedge dx^J=0.$ Hence, the sum is actually over disjoint multi-indices:
  $$
\omega\wedge\tau=\sum_{I,J\text{ disjoint}}(a_Ib_J)dx^I\wedge dx^J,
$$
  which shows that the wedge product of two $C^\infty$ forms is $C^\infty$.

Here, he uses the capitol letters $I$ and $J$ to denote strictly increasing sets of indices $I=(i_1<\dots<i_k)$ and $J=(j_1<\dots<j_l)$ of lengths $k$ and $l$, respectively, from the set of indices $\{1,\dots,n\}$. So, for example $dx^I=dx^{i_1}\wedge\dots\wedge dx^{i_k}$ where $i_1,\dots,i_k\in \{1,\dots,n\}$. 
My question is on the equation in $(\ast)$. Essentially, why does it make sense that the pointwise wedge product of $\omega$ and $\tau$ yields the equation in $(\ast)$? I understand the last half of the text above, and I am even able to use the definition of $\omega\wedge \tau$ given in $(\ast)$ to compute a specific example given in the exercises. But how does one arrive at this equation by defining the wedge product of $\omega$ and $\tau$ pointwise?
 A: I suppose I will answer my own question since I think I've figured it out. It comes down to the fact that the functions $a_I$ and $b_J$ are technically 0-forms. Then we use algebraic properties of the wedge product and look at what those algebraic properties tell us when a 0-form is involved. 
First, by the anticommutativity of $\wedge$, which says that for a $k$-form $f$ and an $\ell$-form $g$
 $$f\wedge g=(-1)^{k\ell}g\wedge f.$$ 
So, for a 0-form $f$ and an $\ell$-form $g$, we get $f\wedge g=(-1)^{0\cdot \ell} g\wedge f=g\wedge f$. 
Second, Tu remarks on page 37 that the wedge product of a 0-form $f$ and an $\ell$-form $\omega$ is actually regular multiplication; that is $f\wedge \omega=f\omega$. So, at a point $p$ we have $(f\wedge\omega)_p=f(p)\omega_p$. 
Therefore, if we assume for a moment that $\omega=a_Idx^I$ and $\tau=b_J dx^J$, then we can view this as $\omega=a_I\wedge dx^I$ and $\tau=b_J\wedge dx^J$. Then
$$
\omega\wedge \tau=a_I\wedge dx^I\wedge b_J\wedge dx^J=a_I\wedge b_J\wedge dx^I\wedge dx^J=(a_Ib_J)\wedge dx^I\wedge dx^J=(a_Ib_J) dx^I\wedge dx^J.
$$
Here's an example:
On Problem 4.3 at the end of the section, we are asked to compute $dx\wedge dy$ where $x=r\cos\theta$ and $y=r\sin\theta$. We get:
$$
dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta
\\
dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta,
$$
and so
$$
dx=\cos\theta dr-r\sin\theta d\theta,
\\
dy=sin\theta dr+r\cos\theta d\theta.
$$
Remember that there is a $\wedge$ between a 0-form and a 1-form; i.e., $\frac{\partial x}{\partial r}dr=\frac{\partial x}{\partial r}\wedge dr$.  Now,
\begin{align*}
dx\wedge dy&= \left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta\right)\wedge\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta\right)\\
&= \left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial r}dr\right)
+\left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial\theta}d\theta\right)
+\left(\frac{\partial x}{\partial\theta}d\theta\wedge\frac{\partial y}{\partial r}dr\right)
+\left(\frac{\partial x}{\partial \theta}d\theta\wedge \frac{\partial y}{\partial\theta}d\theta\right)\\
&= \left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial r}dr\wedge dr\right)
+\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right)
+\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right)
+\left(\frac{\partial x}{\partial \theta} \frac{\partial y}{\partial\theta}d\theta\wedge d\theta\right)\\
&=0+\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right)
+\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right)+0\\
&=r dr\wedge d\theta.
\end{align*}
The first equal sign is by definition; the second is because $\wedge$ is distributive over addition; the third is by the anticommutativity of $\wedge$, where, for example we have $dr\wedge \frac{\partial y}{\partial r}=\frac{\partial y}{\partial r}\wedge dr=\frac{\partial y}{\partial r}dr$ in the first parenthesis; the fourth equal sign is by the fact that
$dr\wedge dr=0=d\theta\wedge d\theta$; the fifth equal sign follows after plugging in the partials and simplifying. 
A: I think another way of looking at the definition of wedge product of differential forms given by Loring Tu in eq. ($*$) is the following paraphrased description. 
Let $\omega$ and $\tau$ be differential $k$-form and $l$-form, respectively. We introduce a differential $(k+l)$-form 
$$\gamma: U \to \bigcup_{p \in U} A_{k+l}\left(T_p^*\left(\mathbb{R}^n\right) \right)$$
such that $\gamma(p) \equiv \gamma_p = \omega_p \wedge \tau_p \in A_{k+l}\left(T_p^*\left(\mathbb{R}^n\right) \right)$. The differential $(k+l)$-form  $\gamma$ is called the wedge product of the differential $k$-form  $\omega$ and $l$-form $\tau$ and is also denoted by $\omega \wedge \tau$.  
