I am asked the following question:
Question: Given that $\Delta f(x) = f(x+1)-f(x)$ and $\Delta_n f(x) = \Delta \left( \Delta_{n-1} f(x) \right)$, prove that for any nth degree polynomial $P(x)$ $$\Delta_{n+1} P(x) = 0 \quad \forall x \in \mathbb{R}$$
I came up with the following solution, although I am not sure it is correct.
My solution It looks like an induction proof would do the trick. Let me start with the base case (although not necessary, I actually tested for two different values of $n$)
For $n=0$: $P(x)=k$
$$\Delta_{n+1} P(x) = \Delta P(x) = P(x+1)-P(x) = k - k = 0$$ (it holds)
For $n=1$: $P(x)=ax+b$
$$\Delta_{n+1} P(x) = \Delta \left( \Delta P(x) \right) = \Delta \left( P(x+1)-P(x) \right) =$$ $$= \Delta \left( a(x+1)+b-ax-b \right) = \Delta \left(a\right) = a-a = 0$$ (it holds)
Inductive hypothesis: Assume $\Delta_{k+1} P(x) = 0$ given $\deg P(x) \leq k$.
Inductive step:
$$\Delta_{(k+1)+1} P(x) = \Delta_{k+2} P(x) = \Delta \left( \underbrace{\Delta_{k+1} P(x)}_\text{$\deg = 0$} \right) = \Delta \left(k \right) = k - k = 0$$
Is the solution above correct?