Let $\varphi : M^n \to N^n$ be a proper smooth map between two connected smooth manifolds. Then $\varphi$ induces a linear map $\varphi^* : H_c^n(N) \simeq \mathbb{R} \to H_c^n(M) \simeq \mathbb{R}$ (where $H_c^n$ is the $n$-th de Rham cohomology group with compact support), so there exists $d \in \mathbb{R}$ so that $\varphi^* : x \mapsto d \cdot x$; $d$ is called the degree of $\varphi$. Moreover, it can be shown that if $\varphi : M \to N$ and $\phi : M\to N$ are smoothly homotopic, then $\varphi^*= \phi^*$ hence $\deg(\varphi)= \deg(\phi)$.
However, every proper smooth map $f : \mathbb{R}^n \to \mathbb{R}^n$ and $g : \mathbb{R}^n \to \mathbb{R}^n$ are smoothly homotopic (thanks to $H(t,x)= tf(x)+(1-t)g(x)$), so every proper smooth map $\mathbb{R}^n \to \mathbb{R}^n$ should have the same degree...
It is probably a silly question, but where does my argument fail?
Edit: Sean Eberhard found a first problem; in fact, to conclude that $\varphi^*=\phi^*$ it is needed the homotopy be proper at any $t$. However, here is another contradiction:
If $P(z)=z^n$ (where $\mathbb{R}^2$ and $\mathbb{C}$ are not distinguised), $\deg(P)=n$ and $P$ is proper if $n \geq 1$. Taking $H(t,z)=tz^m+(1-t)z^n$ with $m>n>1$, $m= \deg(H(1, \cdot))= \deg(H(0,\cdot))=n$, a contradiction.
In the interest of marking this question as answered, I'll just say that properness is important throughout the homotopy. See the comments for discussion.