, 1997

For any function $f(x)$ and any integer $m \ge 0$, define
$$\Delta_m f(x) = \sum_{n=0}^{m} f(x+n) (-1)^n \binom{m}{n}.$$
Claim. Let $Q(x)$ be a polynomial of degree $d$. Then $\Delta_m Q(x) = 0$ for all $m \ge d$.
Proof. This is a standard result about finite differences of polynomials. Indeed, let $Q_0(x) = Q(x)$ and $Q_{k+1}(x) = Q_k(x) - Q_k(x+1)$ for $k \ge 0$. It is easy to show that $Q_k(x) = \Delta_k Q(x)$ for any $k$. Also, it can be proved by induction that $Q_k(x)$ is a polynomial of degree $d-k$ for $k=0, 1, \dots, d$ (essentially because $x^j - (x+1)^j$ has degree $j-1$), and hence $Q_k(x) = 0$ for all $k \ge d$. This proves the claim. $\square$

Consider any function $f(x)$ such that $f(n) = a_n$ for all integer $n \ge 0$. We are given that $\Delta_m f(0) = 0$ for all $m \ge N$ where $N \in \mathbb{Z}^+$. By the Lagrange interpolation formula, construct a polynomial $P$ of degree at most $N-1$ such that $P(n) = f(n)$ for $n = 0, 1, \dots, N-1$. By the claim, we have $\Delta_N P(0) = 0$. Also, we know that $\Delta_N f(0) = 0$. Equating these, we find that
$$\sum_{n=0}^{N} P(n) (-1)^n \binom{N}{n} = \sum_{n=0}^{N} f(n) (-1)^n \binom{N}{n}.$$
Since $P(n) = f(n)$ for $n = 0, 1, \dots, N-1$, this yields $P(N) = f(N)$. Inductively, we obtain $P(n) = f(n) = a_n$ for all nonnegative integer $n$. This completes the proof.