Theorem 4.9. If $f:\mathbb{R}^n \to \mathbb{R}^n$ is differentiable, then $$f^*(hdx^1 \wedge \cdots \wedge dx^n)=(h\circ f)(\det f')dx^1\wedge \cdots \wedge dx^n.$$
I am trying to provide a detailed proof of this theorem, however I am stuck at some step. Here it is my attempt.
Proof. First note that $dx^1 \wedge \cdots \wedge dx^n$ is a $n$-form on $\mathbb{R}^n$, since for $q \in \mathbb{R}^n$ we have $$dx^1 \wedge \cdots \wedge dx^n(q)=dx^1(q) \wedge \cdots \wedge dx^n(q)\in \Omega^n(\mathbb{R}_q^n).$$ Hence $f^*(dx^1 \wedge \cdots \wedge dx^n)$ is a $n$-form in $\mathbb{R}^n.$
Now, according to the Theorem 4-8(3) in Spivak, we have $$f^*(hdx^1\wedge \cdots \wedge dx^n)=(h \circ f)\cdot f^*(dx^1 \wedge \cdots \wedge dx^n).$$ It suffices to show that $$f^*(dx^1 \wedge \cdots \wedge dx^n)=(\det f')dx^1 \wedge \cdots \wedge dx^n.$$
Let $p \in \mathbb{R}^n$ and let $A=(a_{ij})$ be the matrix of $f'(p).$ We need to prove that $$\tag{1} f^*(dx^1 \wedge \cdots \wedge dx^n)(p)=(\det f')dx^1 \wedge \cdots \wedge dx^n(p) $$ right?
Since $dx^1 \wedge \cdots \wedge dx^n(p) \in \Omega^n(\mathbb{R}_p^n),$ then the domain of $dx^1 \wedge \cdots \wedge dx^n(p)$ is $\mathbb{R}_p^n\times \cdots \times \mathbb{R}_p^n.$ In order to prove the equality in (1), we should first prove that the following equality holds:
\begin{align} \tag{2}
f^*(dx^1 \wedge \cdots \wedge dx^n)(p)((e_1)_p, \ldots, (e_n)_p) \\
= \qquad \qquad \qquad \qquad \quad\\
(\det f')dx^1 \wedge \cdots \wedge dx^n(p)((e_1)_p, \ldots, (e_n)_p)
\end{align}
Observe now that \begin{align} f^*(dx^1 \wedge \cdots \wedge dx^n)(p)((e_1)_p, \ldots, (e_n)_p) &= \\ f^*(dx^1 \wedge \cdots \wedge dx^n(f(p)))((e_1)_p, \ldots, (e_n)_p) &= \\ dx^1 \wedge \cdots \wedge dx^n(f(p))\left(f_*\left((e_1)_p\right), \ldots, f_*\left((e_n)_p\right)\right). \end{align}
Now by definition we have $f_*((e_j)_p)=(Df(p)(e_j))_{f(p)}=\left( \sum_{i=1}^n a_{ij}e_i \right)_{f(p)}=\sum_{i=1}^n a_{ij} (e_i)_{f(p)}.$ Then by the previous discussion we now have the equality \begin{align} f^*(dx^1 \wedge \cdots \wedge dx^n)(p)((e_1)_p, \ldots, (e_n)_p) &= \\ dx^1 \wedge \cdots \wedge dx^n(f(p))\left(\sum_{i=1}^n a_{i1} (e_i)_{f(p)}, \ldots,\sum_{i=1}^n a_{in} (e_i)_{f(p)} \right) &= \\ \det (a_{ij}) dx^1 \wedge \cdots \wedge dx^n(f(p)) ((e_1)_{f(p)}, \ldots, (e_n)_{f(p)}). \end{align} (The last equality follows inmediatly from Theorem 4-6).
Now the problem is here: How should I deduce the equality (2) from this last equation? It is not clear at all even if the equality (2) is true. Or did I missed something at some step? Could you help me solving this? Thanks in advance.