The degree of Gauss map

If $M$ is an $2m$-dimensional closed orientable hypersurface in $\mathbb R^{2m+1}$, then we have a Gauss map $G:M\rightarrow S^{2m}$.

I have known from my differential geometry book that deg$G=\frac{1}{2}\chi(M)$ where $\chi(M)$ is the Euler characteristics of $M$. So my question is: is there an easy way to prove this conclusion if we assume the Poincaré–Hopf theorem and why is this argument only applied to the even dimension?

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I guess the point will turn out to be that $\deg G=\frac{\chi(M)}{\chi(S^n)}$, but in odd dimension this is division by zero. – Chris Eagle Oct 19 '12 at 6:48
Interesting thing is, $\chi (M)$ is zero in odd dimension – Hezudao Oct 19 '12 at 18:54