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...

Another way to see it: prove that
$$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\,\,\,,\,\,\text{so}$$
$$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k}+\sum_{k=0}^n(-1)^k\cdot 1^{n-k}\binom{n}{k+1}=$$
$$=(1+(-1))^n-\sum_{k=0}^n(-1)^{k+1}\cdot 1^{n-k-1}\binom{n}{k+1}=0-(1-1)^n+1=1$$
