Proof that $\sum_{i=0}^n 2^i = 2^{n+1} - 1$ $\sum_{i=0}^n 2^i = 2^{n+1} - 1$
I can't seem to find the proof of this. I think it has something to do with combinations and Pascal's triangle. Could someone show me the proof? Thanks

\begin{eqnarray}
2^{n+1} &=& 2^n+2^n \\
&=&2^n + 2^{n-1} + 2^{n-1} \\
&\vdots& \\
&=&  2^n + 2^{n-1} +2^{n-1} + \cdots + 2 +2 \\
&=&  2^n + 2^{n-1} +2^{n-1} + \cdots + 2 +1 + 1
\end{eqnarray}
