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How can I show that $\sum _{n=0}^{p-1} (-1)^n \binom{p}{n+1}=1$ where $p$ is a positive integer?

Is this the rule for this proof?

I could not figure it out even using this.

Probably this may be a well-known result, however I could not find.

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By the binomial theorem, $$ 0=(1-1)^p=\sum_{j=0}^p(-1)^j{p\choose j}$$ Note that the $j=0$ term in the sum is $1$. Therefore $$ 1=-\sum_{j=1}^p(-1)^j{p\choose j}=\sum_{j=1}^p(-1)^{j-1}{p\choose j}$$ and setting $n=j-1$, this becomes $$ 1=\sum_{n=0}^{p-1}(-1)^n{p\choose n+1}$$

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  • $\begingroup$ It needs a simple but smart trick, thanks @carmichael561. $\endgroup$ – Frey Jul 5 '17 at 2:04

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