TITLE: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$
QUESTION [22 upvotes]: I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$.

REPLY [3 votes]: Using the binomial theorem, we find:
$2^n = (1 + 1)^n = \sum\limits_{k=0}^{n} \binom{n}{k} 1^k*1^{n-k} = \sum\limits_{k=0}^{n} \binom{n}{k}$