Compact Manifold with odd dimension has Euler numer 0. (Proof using differential topology)

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So I am trying to prove this, as an excercise for a course in differential topology, the previous part of the excercise was to find $\chi(S^n)$, which is $1+(-1)^n$, it is likely that this fact is meant to use in this part.

I really do not know how to begin, and all proofs I have found, use algebraic topology.

Any help would be appreciated

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HINT: Let $\vec v$ be a "good" vector field on your compact manifold $M$. (So it has isolated zeroes, each with index $\pm 1$, although we really don't need this.) If $p$ is a zero of $\vec v$ with index $k$, what is the index of $-\vec v$ at $p$?