Show that $\Delta^mp(x)= 0 $ when $p(x) \in \mathscr P_n$ and $m\ge n+1$, where $\mathscr P_n$ is the set of polynomial of degree $n$ and $\Delta^m$ is the operator for the $m$-th forward difference.
Here's what I've done:
$$\Delta p(x)= \sum_{i=0}^{1}{1 \choose i}(-1)^ip(x+(k-i)h)= p(x+h)-p(x)$$
We can then rewrite both of these with the substitution $x=x_0+rh$ to Newtons form of interpolating functions: $$p(x) = f(x_0)+\Delta f(x_o){r \choose 1} + \Delta^2f(x_0){r \choose 2}+...+\Delta^nf(x_0){r \choose n}$$ $$p(x+h) = f(x_0)+\Delta f(x_o){r-1 \choose 1} + \Delta^2f(x_0){r-1 \choose 2}+...+\Delta^nf(x_0){r-1 \choose n}$$
Then take $$p(x+h)-p(x)=\frac{rp(x+h)-\left(r-(n-2)\right)p(x)}{r(r-(n-2))}$$ The numerator has degree $n+1$ and the denominator is of degree $2$ therefore the degree of $\Delta p(x)$ is of $n-1$.
This shows that each time we take the forward difference $\Delta$ of $p(x)$ the degree decreases by $1$. Thus if $p(x)\in \mathscr P_n$, $\Delta p(x)\in \mathscr P_{n-1}$. From this we can conclude that $\Delta^{n-1}p(x)$ is linear, $\Delta^{n}p(x)$ is a constant and $\Delta^{n+1}p(x)=0.$
Is this argument true? Not sure if I did the computations correct.
Actually, whatever is the polynomial $p(x) \in \mathcal{P}_n$, you have $p(x+h) - p(x) \in \mathcal{P}_{n-1}$ for $h \in \mathbb{R}$ (interpolation plays no role here): if $p(x) = \sum_{i=1}^n a_i x^i$, then $p(x+h) = \sum_{i=1}^n a_i (x+h)^i$. So,
$$ p(x+h) - p(x) = \sum_{i=1}^n a_i \big( (x+h)^i - x^i \big)$$
Now, in the sum, the only term that could be of degree $n$ is when $i=n$ (all the others have lower degree). But actually, $(x+h)^n - x^n$ is of degree $n-1$, so $p(x+h)-p(x)$ is of degree at most $n-1$.
Then, as you said, each time you apply the forward difference operator, you loose one degree, so in the end, you get $0$.