It is easy to compute that $\Delta f(n)=f(n+1)-f(n)=n+1,\;\Delta^2 f(n)=\Delta f(n+1)-\Delta f(n)=(n+2)-(n+1)=1$ and $\Delta^k f(n)=0$ for $k\geq 3$. Can we conclude that $f$ is polynomial of degree $2$? I know that in this given example $f$ is polynomial of degree $2$ but what about general case?
If $f$ is function for which $\Delta^{k+1} f(n)= \Delta^k f(n+1)-\Delta^k f(n)=0$ for $k\geq 3$ then $f$ is polynomial of degree $2$?
Yes, and this is a result that holds generally (except it is of degree at most 2). We can even prove it. First, the more general statement:
We proceed by induction. As base case we have a function such that $\Delta f(n) = 0$. What does this mean? It means the value of $f$ never changes. So $f$ must be a constant, which is a polynomial of degree $0$ (the zero polynomial itself has degree $0$ in this case).
Now for induction, assume we have proven the above statement for all values of $k$ up to $\ell - 1$, and want to prove that it holds for $k = \ell$ as well. So we take a function $f$ such that $\Delta^{\ell + 1}f(n) = 0$. That means, by the same argument as the base case, that $\Delta^\ell f(n)= c$ is a constant. Now consider the polynomial function given by $g(n) = \frac{c}{\ell!}n^{\ell}$. We have $\Delta^\ell g(n) = c$. So this means we have $\Delta^\ell(f-g)(n) = 0$. By the induction hypothesis, this means that $f-g$ is a polynomial of degree at most $\ell - 1$. And since $g$ is a polynomial (either the zero polynomial or a polynomial of degree $\ell$) this means that $f$ must be a polynomial of degree at most $\ell$.