Why is the following implication true?
${d\over dt}\vec v(t)=\vec n\times \vec v(t) \implies \vec n\cdot\vec v(t)=\vec n\cdot \vec v(0)$
I think the result can be obtained by expressing $\vec v(t)$ in a Taylor series expansion in $t$. And from $\displaystyle {d\over dt}\vec v(t)=\vec n\times \vec v(t) $, we see that $\vec n\cdot \vec v^{(n)}(0)=0$ due to the $\times \vec n$ bit.
Is there a more direct way?