Alternating sum of binomial coefficients 
Calculate the sum:
$$ \sum_{k=0}^n (-1)^k {n+1\choose k+1} $$

I don't know if I'm so tired or what, but I can't calculate this sum. The result is supposed to be $1$ but I always get something else...
 A: HINT


*

*$\displaystyle (1-1)^{n+1} = \sum_{k = 0}^{n+1} {n \choose k}1^k(-1)^{n-k}$

*What terms are or aren't in your sum that are in the one above?

A: Using the binomial theorem we have:
$$ (1 + (-1))^{n+1} = {{n+1} \choose 0} (-1)^0 + {{n+1} \choose 1} (-1)^1 + \ldots + {{n+1} \choose {n+1}} (-1)^{n+1}.$$
"Divide" by ${-1}$ to get:
$$ - (1 - 1)^{n+1} = -{{n+1} \choose 0} + \color{blue}{{{n+1} \choose 1} (-1)^0 + \ldots + {{n+1} \choose {n+1}} (-1)^{n}}.$$
This pretty much solves it.
A: Another way to see it: prove that
$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\,\,\,,\,\,\text{so}$$
$$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k}+\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k+1}=$$
$$=(1+(-1))^n-\sum_{k=0}^n(-1)^{k+1}\cdot 1^{n-k-1}\binom{n}{k+1}=0-(1-1)^n+1=1$$
A: $$\sum_{k=0}^n (-1)^k {n+1\choose k+1}=-\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}= $$
$$=-\left(\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}\right)=$$
$$=-\left(\sum_{j=1}^{n+1} (-1)^{j} {n+1\choose j}+(-1)^{0} {n+1\choose 0}-(-1)^{0} {n+1\choose 0}\right)= $$
$$=-\left(\sum_{j=0}^{n+1} (-1)^{j} {n+1\choose j}-(-1)^{0} {n+1\choose 0}\right)=-(1-1)^{n+1}+1=1$$
