I'm trying to prove a statement suggested by @reuns from here.
Let $E$ be a vector space over field $\mathbb{K}$. To simplify notation, we write ${\operatorname{G}f}_n$ for $(\operatorname{G}(f))_n$ where $\operatorname{G} \in \operatorname{End}\left(E^{\mathbb{N}}\right)$ and $f \in E^{\mathbb{N}}$. On $E^{\mathbb{N}}$ define difference operator $\triangle$ by $$\triangle f_{n} :=f_{n+1}-f_{n}$$
Theorem: Assume $\mathbb K$ is a field containing $\mathbb N$, $f \in \mathbb{K}^\mathbb{N}$ such that ${\triangle^k f}$ is constant, and there exists a polynomial $p \in \mathbb{K}_k[X]$ such that $f_n = p(n)$ for all $n \le k$. Then $f_n = p(n)$ for all $n \in \mathbb N$.
Could you please verify my attempt? Any suggestion is greatly appreciated.
My attempt:
Lemma: For $f \in \mathbb K^\mathbb N$, we have $$f_{n+k}=\sum_{j=0}^{k} {k \choose j} \triangle^{j} f_{n} \quad (\star) \text{ and } \triangle^{k} f_{n}=\sum_{j=0}^{k}(-1)^{k-j} {k \choose j} f_{n+j} \quad (\star \star)$$
We define the sequence $g_n$ by $g_n := p(n)$ for all $n \in \mathbb N$. We will prove by induction that $$f_{k+n} = g_{k+n}, \quad n \in \mathbb N$$
The statement clearly holds for $n = 0$. Assume that $f_{k+i} = g_{k+i}$ for all $i \le n$. Then $f_t = g_t$ for all $t \le k+n$. Notice that because $\triangle^{k} f$ is constant and $p$ has degree $k$, $\triangle^{k+1} f_{n} = \triangle^{k+1} g_{n} = 0$ for all $n \in \mathbb N$.
From $(\star)$, we have $$\begin{aligned} f_{k + (n+1)} &= f_{n + (k+1)} \\ &= \sum_{j=0}^{k+1} {k+1 \choose j} \triangle^{j} f_{n} \\ &= \left (\sum_{j=0}^{k} {k+1 \choose j} \triangle^{j} f_{n} \right ) + {k+1 \choose k+1} \triangle^{k+1} f_{n}\\ &= \sum_{j=0}^{k} {k+1 \choose j} \triangle^{j} f_{n} \end{aligned}$$
Similarly, $$\begin{aligned}g_{k + (n+1)} &= \left (\sum_{j=0}^{k} {k+1 \choose j} \triangle^{j} g_{n} \right ) + {k+1 \choose k+1} \triangle^{k+1} g_{n} \\ &= \sum_{j=0}^{k} {k+1 \choose j} \triangle^{j} g_{n}\end{aligned}$$
From $(\star \star)$, we have $\triangle^{j} f_{n} = \sum_{i=0}^{j}(-1)^{j-i} {j \choose i} f_{n+i}$. On the other hand, our inductive hypothesis implies $f_{n+i} = g_{n+i}$ for all $i \le j \le k$. Thus $\triangle^{j} f_{n} = \triangle^{j} g_{n}$ for all $j \le k$. As such, $f_{k + (n+1)} = g_{k + (n+1)}$. This completes the proof.