I want to know if I can get some help with this proof. I tried, but I just cannot seem to get $2^{k}$. It states that,     For $k \in \mathbb{Z}_{\ge 0}$, $$\sum^{k}_{m=0}\binom{k}{m} = 2^k$$   Thank You.
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 $2^n$.
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