The Poincaré-Hopf theorem states that for a smooth compact $m$-manifold $M$ without boundary and a vector field $X\in\operatorname{Vect}(M)$ of $M$ with only isolated zeroes we have the equality $\sum_{\substack{p\in M\\X(p)=0}}\iota(p,X)=\chi(M)$ where $\iota(p,X)$ denotes the index of $X$ at $p$ and $\chi(M)$ denotes the Euler characteristic of $M$.
Let $m$ be even and $M\subset\mathbb{R}^{m+1}$ be a $m$-dimensional smooth compact submanifold without boundary and denote by $\nu:M\to S^m$ its Gauss map. How can I deduce from the Poincaré-Hopf theorem that the Brouwer degree of $\nu$ is equal to half the Euler characteristic of $M$ i.e. $\deg(\nu)=\frac{1}{2}\chi(M)?$
After many repeated (unsuccessful) tries I was hoping hat someone else might shed some light onto this...