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...

$$\sum_{k=0}^n (-1)^k {n+1\choose k+1}=-\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}= $$
$$=-\left(\sum_{k=0}^n (-1)^{k+1} {n+1\choose k+1}\right)=$$
$$=-\left(\sum_{j=1}^{n+1} (-1)^{j} {n+1\choose j}+(-1)^{0} {n+1\choose 0}-(-1)^{0} {n+1\choose 0}\right)= $$
$$=-\left(\sum_{j=0}^{n+1} (-1)^{j} {n+1\choose j}-(-1)^{0} {n+1\choose 0}\right)=-(1-1)^{n+1}+1=1$$
