Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$

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 $$g(n) = \sum_{i=0}^n \binom{n}{i} = 2^n.$$

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.

Well, here is one.

$$\sum_{i=0}^n \binom{n}{i}=2^n$$ $$\sum_{i=0}^n \binom{n}{i}+\sum_{i=0}^n \binom{n}{i}=2^{n+1}$$ $$\binom{n}{0}+\left [ \binom{n}{0}+\binom{n}{1} \right ]+...+\left [ \binom{n}{n-1}+\binom{n}{n}\right ]+\binom{n}{n}=2^{n+1}$$ $$\sum_{i=0}^{n+1} \binom{n+1}{i}=2^{n+1}$$