Differences of polynomials vanish beyond the degree

Remarks

The key step is that $\Delta_h$ lowers the degree: for $m \ge 1$, $(x + h)^m - x^m = \sum_{j<m} \binom{m}{j} h^{m-j} x^j$ has degree less than $m$, and $\Delta_h$ sends constants to $0$. Induction on $n$ then gives the claim. The same computation shows more: if $p$ has degree $d$ and leading coefficient $a$, then $\Delta_h^d p$ is the constant $d!\, a\, h^d$. Example: for $p(x) = x^3$ and $h = 1$, the third differences of $0, 1, 8, 27, 64$ are all $6 = 3!$, and the fourth differences are $0$. Over $\R$ there is a converse: a function $f \colon \Z \to \R$ with $\Delta^n f = 0$ agrees on $\Z$ with a polynomial of degree less than $n$.

Going deeper

For a student

The fact gives a practical test: if a table of values has constant $d$-th differences, the data fit a polynomial of degree $d$. With the expansion of iterated differences in shifts it yields identities such as $\sum_{k=0}^{n} (-1)^k \binom{n}{k} k^m = 0$ for $0 \le m < n$. Combined with Newton's forward difference formula, it shows that a polynomial of degree $d$ is determined by its values at $d + 1$ equally spaced points. It is then given by a finite Newton series.