Let $M,N$ be $n$-dimensional oriented smooth manfiolds without boundary and $\omega$ a compactly supported $n$-form on $N$. If $f \colon M \rightarrow N$ is an orientation-preserving diffeomorphism, then $$ \tag{1} \label{a} \int_{N}\omega = \int_{M}f^{\ast}\omega. $$
I want to prove (\ref{a}) using the change of variables formula in $\mathbb{R}^n$, which says that if $\psi \colon U \rightarrow V$ is an orientation-preserving diffeomorphism between open subsets of $\mathbb{R}^n$, then $$ \tag{2} \label{b} \int_{V}\eta = \int_{U}\psi^{\ast}\eta, $$ where $\eta$ is an $n$-form having compact support in $V$.
Try of proof: W.l.o.g. we can assume (otherwise take a partition of unity) that there is an orientation-preserving chart $\varphi \colon U \rightarrow O$ such that $\operatorname{supp} \omega \subseteq O$, where $O \subseteq N$ is open. Therefore by definition of the integral $$ \tag{3} \label{c} \int_{N}\omega = \int_{U} \varphi^{\ast}\omega. $$ On the other hand, $\operatorname{supp}f^{\ast}\omega \subset f^{-1}(O)$ and $f^{-1} \circ \varphi$ is a chart for $M$ with $\operatorname{supp}f^{\ast}\omega \subset \operatorname{im} f^{-1} \circ \varphi$. Therefore $$ \tag{4} \label{d} \int_{M}f^{\ast}\omega = \int_{U}(f^{-1} \circ \varphi)^{\ast}f^{\ast}\omega. $$ At this point I would be tempted to write $(f^{-1} \circ \varphi)^{\ast}f^{\ast}\omega = \varphi^{\ast}\omega$ and say the integrals (\ref{c}) and (\ref{d}) are equal, but I think something is wrong in this argument. However, how can I rewrite (\ref{a}) to reduce it to the case (\ref{b})?