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

Recall the relation $\displaystyle{n+1\choose i}={n\choose i}+{n\choose i-1}$, valid for every $1\leqslant i\leqslant n$. Hence,
$$
\sum_{i=0}^{n+1}{n+1\choose i}=1+\sum_{i=1}^n\left[{n\choose i}+{n\choose i-1}\right]+1,
$$
that is,
$$
\sum_{i=0}^{n+1}{n+1\choose i}=1+\sum_{i=1}^n{n\choose i}+\sum_{i=0}^{n-1}{n\choose i}+1=2\sum_{i=0}^n{n\choose i}.
$$
The initial value $\displaystyle\sum\limits_{i=0}^0{0\choose i}=1$ completes the recursion.
