Spivak Calculus on Manifolds - Theorem 4-10

I am posting my own answer, using the notation and theorems in Spivak, just to clarify my understanding.

Suppose $\omega$ is a $0$-form on $\mathbb{R}^m$, that is, $\omega : \mathbb{R}^m \to \mathbb{R}$ is a $C^\infty$ function. Then, $$ d\omega = \sum_{\alpha=1}^m D_\alpha \omega \cdot dx^\alpha \qquad \text{(by Theorem 4-7)} $$ is a $1$-form on $\mathbb{R}^m$. Let $\, f : \mathbb{R}^n \to \mathbb{R}^m$ be a differentiable function. $\, f^*(d\omega)$ is thus a $1$-form on $\mathbb{R}^n$. Let $p \in \mathbb{R}^n$. Then, $$ \begin{align*} f^*(d\omega)(p) &= f^* \left( \sum_{\alpha=1}^m D_\alpha \omega \cdot dx^\alpha \right) (p) \\ &= \sum_{\alpha=1}^m f^* (D_\alpha \omega \cdot dx^\alpha) (p) &&\text{(by Theorem 4-8(2))}\\ &= \sum_{\alpha=1}^m [(D_\alpha (\omega) \circ f) \cdot f^*(dx^\alpha)](p) &&\text{(by Theorem 4-8(3))}\\ &= \sum_{\alpha=1}^m \left[ (D_\alpha(\omega) \circ f) \cdot \sum_{\beta=1}^n D_\beta \;\! f^\alpha \cdot dx^\beta \right](p) &&\text{(by Theorem 4-8(1))}\\ &= \sum_{\beta=1}^n \left[ \sum_{\alpha=1}^m D_\alpha \omega(f(p)) \cdot D_\beta \;\! f^\alpha(p) \right] dx^\beta(p)\\ &= \sum_{\beta=1}^n D_\beta(\omega \circ f)(p) \cdot dx^\beta(p) &&\text{(by Theorem 2-9)}\\ &= \left(\sum_{\beta=1}^n D_\beta(\omega \circ f) \cdot dx^\beta \right) (p). \end{align*} $$ Hence, $$ f^*(d\omega) = \sum_{\beta=1}^n D_\beta(\omega \circ f) \cdot dx^\beta. $$ On the other hand, $\, f^* \omega$ is a $0$-form on $\mathbb{R}^n$ given by $$ f^* \omega(p) = \omega(f(p)) = \omega \circ f(p) $$ for all $p \in \mathbb{R}^n$. That is, $$ f^* \omega = \omega \circ f. $$ Therefore, $d(f^* \omega)$ is a $1$-form on $\mathbb{R}^n$, and we have $$ \begin{align*} d(f^* \omega) &= \sum_{\beta=1}^n D_\beta(f^* \omega) \cdot dx^\beta &&\text{(by Theorem 4-7)}\\ &= \sum_{\beta=1}^n D_\beta(\omega \circ f) \cdot dx^\beta. \end{align*} $$ Thus, if $\omega$ is a $0$-form on $\mathbb{R}^m$ and $\, f : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable, then $$ f^*(d\omega)=d(f^*\omega). $$