$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$ I am reading Guillemin and Pollack's Differential Topology. For the proof on Page 164, I was not able to get through the last step.

$$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$$
Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

According to Daniel Robert-Nicoud's ansewr to $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, what is $\omega[f(x)]$?
Locally, differential form can be written as
$$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}$$
with $\alpha$ a smooth function. Then
$$f^*\omega_\alpha(x) = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}].$$
We write
$$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p},$$
$$\theta_\beta(y) = \beta(y)dx^{j_1}\wedge\ldots\wedge dx^{j_q}.$$
Hence,
$$\omega \wedge \theta = (\alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}) \wedge (\beta(y)dx^{j_1}\wedge\ldots\wedge dx^{j_q})$$
Following James S. Cook's very brilliant answer Pullback expanded form.
$$\omega \wedge \theta = \alpha(y) \beta(y) \sigma(I,J) dx^{k_1}\wedge\ldots\wedge dx^{k_{p+q}}$$
Rename $\gamma(y) = \alpha(y) \beta(y) \sigma(I,J)$, we get
$$f^*(w \wedge \theta) =  (df_x)^* [(\gamma \circ f)(x) dx^{k_1}\wedge\ldots\wedge dx^{k_{p+q}}].$$
But $$(f^*w) \wedge (f^* \theta) = ((df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}])\wedge((df_x)^*[(\beta\circ f)(x)dx^{j_1}\wedge\ldots\wedge dx^{j_q}])$$

So here I got stuck - I don't really know how to move around $\alpha, \beta$ under $df^*$, to get close to the left hand side expression.

Thank you!~
 A: I read your approach but I am giving another approach which might be helpful. Lets start with the definition. If $f:X\rightarrow Y$ is a map and $\omega$ is a $p$ form in $Y$ then its pull back is defined as $f^*(\omega)v=\omega (f_*v).,$ where $f_*v=v(f),$ for any $v\in T_pX.$ Therefore Expand (assuming $\omega$ is a p form and  $\theta$ is a q form) the term (where $v_i$and $w_i$ are in $T_pX$ ) $f^*(\omega\wedge \theta)(v_1,\cdots ,v_p,w_1,\cdots ,w_q)=(\omega\wedge \theta)(f_*(v_1),\cdots ,f_*(v_p),f_*(w_1),\cdots ,f_*(w_q)).$ using the summation formula for wedge product. Which will be same as $\omega(f_∗(v_1),⋯,f_∗(v_p))\wedge \theta(f_*(w_1),⋯,f_∗(w_q)).$ Which is (by definition) same as $f^*(\omega)\wedge f^*(\theta).$ 
A: By definition,
$$f^*(w \wedge \theta) = (df_x)^*(\omega\wedge\theta)[f(x)].$$

Page 162: If $\omega$ is a $p$-form and $\theta$ is a $q$-form, then $p+q$ form
  $$(\omega \wedge \theta) (x) = \omega(x) \wedge \theta(x).$$

Therefore,
$$(df_x)^*(\omega\wedge\theta)[f(x)] = (df_x)^*\omega[f(x)] \wedge \theta[f(x)].$$

Page 159: $A^*$ is linear and that
  $$A^*(T \wedge S) = A^*T \wedge A^*S.$$

Therefore,
$$(df_x)^*\omega[f(x)] \wedge \theta[f(x)] = (df_x)^*\omega[f(x)] \wedge (df_x)^*\theta[f(x)].$$
That is exactly
$$(f^*w) \wedge (f^* \theta).$$
