Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$ 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 $2^n$.
 A: You could use exponential generating functions to prove this identity.
$$\begin{eqnarray}\sum_{n\ge0}2^n\frac{x^n}{n!}&=&\sum_{n\ge0}\frac{(2x)^n}{n!}\\&=&e^{2x}\\&=&e^xe^x\\&=&\sum_{i\ge0}\frac{x^i}{i!}\sum_{j\ge0}\frac{x^j}{j!}\\&=&\sum_{n\ge0}\sum_{i=0}^{n}\frac{x^i}{i!}\frac{x^{n-i}}{(n-i)!}\\&=&\sum_{n\ge0}\sum_{i=0}^n\binom{n}{i}\frac{x^n}{n!}\end{eqnarray}
$$
Now, comparing the coefficients of $x^n$ for ${n\ge0}$ gives the identity.
A: Using the binomial theorem, we find:
$2^n = (1 + 1)^n = \sum\limits_{k=0}^{n} \binom{n}{k} 1^k*1^{n-k} = \sum\limits_{k=0}^{n} \binom{n}{k}$
A: Here's one.  Let $g(n) = \sum \limits_{i=0}^n \binom{n}{i}$.  Then
$$g(n+1) - g(n) = \sum_{i=0}^{n+1} \binom{n+1}{i} - \sum_{i=0}^n \binom{n}{i} = \sum_{i=0}^{n+1} \left(\binom{n+1}{i} - \binom{n}{i}\right) = \sum_{i=0}^{n+1} \binom{n}{i-1} $$
$$= \sum_{i=0}^n \binom{n}{i} = g(n).$$
Here, we use the fact that $\binom{n}{n+1} = \binom{n}{-1} = 0$, as well as the binomial recurrence $\binom{n+1}{i} = \binom{n}{i} + \binom{n}{i-1}$.
Thus we have $g(n+1) = 2g(n)$, with $g(0) = 1$.  Since $g(n)$ doubles each time $n$ is incremented by 1, we must have 
$$g(n) = \sum_{i=0}^n \binom{n}{i} = 2^n.$$
A: Simply use the binomial formula.
$$(a + b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n - k}$$
With $a = b = 1$ you have your result.
A: Well, here is one. 
$$\sum_{i=0}^n \binom{n}{i}=2^n$$
$$\sum_{i=0}^n \binom{n}{i}+\sum_{i=0}^n \binom{n}{i}=2^{n+1}$$
$$\binom{n}{0}+\left [ \binom{n}{0}+\binom{n}{1} \right ]+...+\left [ \binom{n}{n-1}+\binom{n}{n}\right ]+\binom{n}{n}=2^{n+1}$$
$$\sum_{i=0}^{n+1} \binom{n+1}{i}=2^{n+1}$$
