Sum from $0$ to $n$ of $ n \choose i $? 
Possible Duplicate:
Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$
Evaluation $\sum\limits_{k=0}^n \binom{n}{k}$

Is there a simple proof for this equality:
$$\sum_0^n  {n \choose i}  = 2^n$$
thanks and sorry I forgot the basics

The standard combinatorial proof is that


*

*The LHS counts the number of ways to choose $0$, $1$, $2, \ldots , $ or $ n$ things from a total of $n$ objects.

*The RHS counts the number of ways to go through each of $n$ objects and mark them as "choose" or "don't choose".


With a little thought, these are equal.
An algebraic proof has been posted by Siminore.
