I would greatly appreciate any hint for proving the following.

Question: Let $f:[0, 1] \to {\bf R}$. Can it be proved that if $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f ]=0$ for all $m=1,2, 3,\dots$, then $f$ is necessarily a polynomial of degree less than or equal to $N$?

(Here $[0, 1/(N+m),\dots, (N+m)/(N+m) ; f]$ denotes the divided difference of $f$ on the nodes $0, 1/(N+m), \dots, 1$.)

Remark: In order to make it easier to obtain a proof, it can be supposed that $f$ satisfies some smoothness properties, as for example to be infinitely differentiable or analytic on $[0, 1]$.

Thank you,

G