What is the result of sum $\sum\limits_{i=0}^n 2^i$

Let $S = 2^0 + 2^1 + 2^2 + \cdots + 2^{n}$.

Then $2S = 2^1 + 2^2 + 2^3 + \cdots + 2^{n} + 2^{n+1}$.

Then $$\begin{align*} S = 2S - S &= & & 2^1 &+& 2^2 & + & 2^3 & + & 2^4 &+&\cdots &+& 2^{n} &+& 2^{n+1}\\ && -2^0 -& 2^1 & - & 2^2 & - & 2^3 & - & 2^4 & - & \cdots & - & 2^n \end{align*}$$ How much is that?

Might be overkill, but there is a well known identity for sums of the form $\sum_{i=0}^n x^i$ where $x$ is not 1.

$$\sum_{i = 0}^n x^i = \frac{1- x^{n+1}}{1-x}$$

Now plug in 2 and you have what you seek.

This identity can easily be proven by induction.

I thought I might post a little more elaborate version of Henning's hint (see his comment). $$\begin{align} 1&=2^0\\ 10&=2^1\\ 100&=2^2\\ 1000&=2^3\\ \vdots&=\vdots\\ 10\dots0&=2^n\\ \hline 11\dots1&=2^0+2^1+\dots+2^n\\ 1&=1\\ \hline 100\dots0&=2^0+2^1+\dots+2^n+1=2^{n+1} \end{align}$$ Hence $2^0+2^1+\dots+2^n=2^{n+1}-1$