Possible Duplicate:
Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$
I am trying to prove $\sum \limits_{i=0}^n \binom{n}{i} = 2^n$ by induction. I've been all over the net looking for a solution because I just don't understand how to go about it. I know the binomial theorem is involved somehow. I found this site which seems to explain it pretty well but I just cant wrap my head around the method.
Can you show me the algebra for this, I don't understand how they get were they are going. How do you algebra $\binom{k+1}{\textrm{anything}}$ into $\binom{k}{\textrm{anything}}$? I just can't make sense of it.
So assume that $\sum \binom{k}{i} = 2^k$.
$\sum \binom{k+1}{i} = \binom{k+1}{0} +\binom{k+1}{1} + \cdots +\binom{k+1}{k} + \binom{k+1}{k+1} ,$ the right side should boil down to $2^{n+1}$.
I'm sure after I'm shown this I will still have questions, but I'm just hoping it will make more sense. Thank you.