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.

I don't know what you mean by "algebraic".  Notice that if $N$ is $0$, we have the empty set, which has exactly one subset, namely itself.  That's a basis for a proof by mathematical induction.
For the inductive step, suppose a set with $N$ elements has $2^N$ subsets, and consider a set of $N+1$ elements that results from adding one additional element called $x$ to the set.  All of the $2^N$ subsets of our original set of $2^N$ elements are also subsets of our newly enlarged set that contains $x$.  In addition, for each such set $S$, the set $S\cup\{x\}$ is a subset of our enlarged set.  So we have our original $2^N$ subsets plus $2^N$ new subsets---the ones that contain $\{x\}$.  The number of subsets of the enlarged set is thus $2^N + 2^N$.
Now for an "algebraic" part of the arugment: $2^N + 2^N = 2^{N+1}$.
