summation - Prove: $sum_{x=0}^{n} (-1)^x {n choose x} = 0$

Is there a quick, fancy, way of proving sums such as this?

Prove that:
$$\sum_{x=0}^{n} (-1)^x {n \choose x} = 0$$

A recent homework assignment I turned in had a couple problems similar to the above. For the most part, I used a proof by induction to solve them. They take forever to write up that way and I was wondering if there were any manipulations possible to speed these types of problems up.

Any suggestions would be greatly appreciated!

Answer

$$(1-1)^n=\sum_{x=0}^n {n\choose x}(-1)^x=0$$

Conisder the following expansion $$(x-a)^n=\sum_{i=0}^n{n \choose i}x^{n-i}(-a)^i$$ . . In your question $x =1$ and $a=1$.