This might be a bit silly question but I haven't find direct reference.
Let $\Omega$ be open, bounded and connected in $\mathbb{R}^n$. Assume that $f:\overline{\Omega}\rightarrow \mathbb{R}^n$ is a diffeomorphism. Since $f:\overline{\Omega}\rightarrow f(\overline{\Omega})$ is a bijection, every $y=f(x)$ has only one solution in $\Omega$, i.e. $ \{ f^{-1}(y_1)\}=\{x_1\}$. Due to inverse function theorem $\det f'(x_1) \ne 0$, for all $x_1 \in \Omega$.
Set $y\notin \partial \Omega$ and calculate the degree
$$\deg( f,\Omega,y) = \sum_{x\in ^{-1}(y)} sign \det f'(x)= sign \det f'(x_1) =±1. $$
Can we find the mapping degree of a diffeomorphism in this simply way?
Yes you can. It is usually also not difficult to compute if the map is orientation preserving or reversing.