I have looked at proofs for this online but I don't think I've grasped them well enough. Upon trying to prove this myself, I've found that the nth difference=$$\sum_{r=0}^{n}{n \choose r}(\sum_{p=0}^{n}a_p(c-r)^{n-p})*(-1)^r$$
Where f(X) is a polynomial and f(c-r)=$$\sum_{p=0}^{n}a_p(c-r)^{n-p}$$ I've tried proving this, but I don't seem to be getting anywhere. Also, how do I show that $$(-1)^n*\sum_{p=1}^{n}{n \choose p}p^n(-1)^p$$=n! ?
I think I've managed to do it. To prove the first expression in my main post we'll need to prove two statements.
Statement 1: $$\sum_{p=1}^{n}{n \choose p}p^m(-1)^p=0, m\in\mathbb{N}, 1\le m \lt n$$ Proof: Let m=1, n>1. Then, $\sum_{p=1}^{n}{n \choose p}p^1(-1)^p=\sum_{p=1}^{n}\frac{n}{p}{n-1 \choose p-1}(-1)^pp=-n\sum_{p=0}^{n-1}{n-1 \choose p}(-1)^p=0$ (use the binomial theorem). Assume it holds for m=k, where n>k. Now if m=k+1 and n>k+1, then $\sum_{p=1}^{n}{n \choose p}p^{k+1}(-1)^p=\sum_{p=1}^{n}\frac{n}{p}{n-1 \choose p-1}(-1)^pp^{k+1}=-n\sum_{p=0}^{n-1}{n-1 \choose p}(-1)^p(p+1)^k=-n[(\sum_{p=0}^{n-1}{n-1 \choose p}(-1)^pp^k)+......]$ (The powers of p decrease by one). Then : $-n[(\sum_{p=1}^{n-1}{n-1 \choose p}(-1)^pp^k)+......]$ We assumed that n>k+1 thus n-1>k. Therefore n-1>k-a for some natural number a. By our inductive hypothesis, the theorem we're trying to prove is valid for any natural number k less than the upper bound n. So the expression turns into: -n[0+0+0+0.....+0]=0.
Statement 2: $$ \sum_{p=1}^{n}{n \choose p}p^n(-1)^p=\frac{n!}{(-1)^n}$$ This holds when n=1. Assume that it works when n=k. Now when you're going to try to show that it works for n=k+1, you'll get this: $$-(k+1)[(\sum_{p=1}^{k}{k \choose p}p^k(-1)^p)+.....]$$ The powers of p are going to decrease by 1. Using our inductive hypothesis and statement 1, it is now easy to say that it equals $\frac{(k+1)!}{(-1)^{k+1}.
Now by expanding the first expression of the main post, using the distributive property and using these 2 statements, it is possible to show that the first expression is true.
I'm sorry for being very brief as I'm typing on a phone. If anyone finds a flaw, please let me know.