Let n>=2 and p(x)=x^n + a1*x^(n-1) + ... + an .
Prove that to equation p(x)=sin(x) have at most n roots.
Between each two zeros of there will be a zero of . So can have at most one more zero than . Then between each two zeros of there will be a zero of , and so on.
Now look at what happens when you differentiate n times.