Prove 2^n possibly with the binomial theorem

[gap0063]

Prove for all n\inN
2ⁿ= (\stackrel{n}{0})+(\stackrel{n}{1})+...+(\stackrel{n}{n})

So I used mathematical induction

base case: n=0 so 2⁰=1 and (\stackrel{0}{0})=1

induction step: Let n\inN be given, assume as induction hypothesis that 2ⁿ= (\stackrel{n}{0})+(\stackrel{n}{1})+...+(\stackrel{n}{n})

I think I'm trying to prove 2ⁿ⁺¹= (\stackrel{n+1}{0})+(\stackrel{n+1}{1})+...+(\stackrel{n+1}{n})


but I don't know how to apply the binomial theorem (if I'm even supposed to!)

 

[Office_Shredder]

This is really a combinatorics problem.

Think about how many ways there are to pick a subset out of a set of n elements.

 

[gap0063]

This is really a combinatorics problem.

Think about how many ways there are to pick a subset out of a set of n elements.

I don't understand... :/

 

[HallsofIvy]

Yes, use the binomial theorem! Let x= y= 1 in (x+ y)^n.