If $P$ is a polynomial of degree at most $N$, we have $$\sum_{n=0}^{N+1}P(n)(-1)^n\binom{N+1}{n}=0$$
I tried using pascal's identity $\binom{N+1}{n}=\binom{N}{n}+\binom{N}{n-1}$. I think that differences of polynomials might work.
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Your idea is correct. Just expand using Pascal's identity and use induction should do. Notice $P(n)-P(n+1)$ is a polynomial in $n$ of degree at most $N-1$. – August Liu Nov 04 '20 at 07:25
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You may see: https://math.stackexchange.com/questions/3250397/summing-s-n-m-sum-k-1n-1k-km-n-choose-k-for-mn-n – Z Ahmed Nov 04 '20 at 08:25