I am reading Mathematical Analysis by Andrew Browder and am confused in his calculating the meaning of $\mathbf{f^\ast},$ the "induced mapping" or pullback of $\mathbf{f^\ast}.$ Here's his definition:
Suppose that $V$ is an open set in $\mathbb{R}^m$, $U$ is an open set in $\mathbb{R}^n,$ and $\mathbf{f}$ is a smooth map of $V$ into $U$. Then $\mathbf{f}$ induces a map $\mathbf{f^{\ast}}: \Omega^r(U) \rightarrow \Omega^r(V)$ defined by
$$(\mathbf{f^{\ast}}\omega)_{\mathbf{p}}(\mathbf{v}_1, \ldots, \mathbf{v}_r)=\omega_{\mathbf{f(p)}}(\mathbf{f'(p)v}_1, \ldots \mathbf{f'(p)v}_r).$$ Then he writes $$(\mathbf{f^{\ast}}dx^I)_{\mathbf{p}}(\mathbf{v}_1, \ldots, \mathbf{v}_r)=dx^I(\mathbf{f'(p)v}_1, \ldots \mathbf{f'(p)v}_r)={(df^{i_1} \wedge \ldots \wedge df^{i_r})}_{\mathbf{p}}(\mathbf{v}_1 \ldots \mathbf{v}_r).$$ Why does the second equality hold?