Algebraic proof that collection of all subsets of a set (power set) of $N$ elements has $2^N$ elements In other words, is there an algebraic proof showing that $\sum_{k=0}^{N} {N\choose k} = 2^N$? I've been trying to do it some some time now, but I can't seem to figure it out.

Another approach is identifying the powerset $\mathcal P \, X$ of a set $X$ with the set of functions $X \to 2$ (that is to say with the set of idicator functions of the subsets). Of course this is only useful if you have any previous results on cardinalities of sets of functions between (finite) sets.
