TITLE: Proving $\sum_{k=0}^n(-1)^k\binom nk=0$
QUESTION [23 upvotes]: Show that
$$\sum_{k=0}^n(-1)^k\binom nk=0$$
So for odd $n$ we have an even number of terms. So $\binom nk=\binom n{n-k}$ which have opposite signs. Thus the sum is 0.
For even $n$ we have that
$$\sum_{k=0}^n(-1)^k\binom nk= \binom n0+\sum_{k=1}^{n-1}(-1)^k\binom nk+\binom nn$$
Now
$$\sum_{k=1}^{n-1}(-1)^k\binom nk= \sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right]$$
What would that sum be in the square brackets?

REPLY [4 votes]: Another approach:
For the difference operator $\Delta f (m) = f(m+1)-f(m)$, the $n$th iteration is 
$$\Delta^n f(m)=\sum_{k}(-1)^{n-k}\binom nk f(m+k)$$
Clearly, your sum is the $n$th difference of the constant function $f\equiv 1$, so of course it is zero for $n\ge 1$ because already $f(m+1)-f(m) =0 $.