Prove 2^n possibly with the binomial theorem

[gap0063]

Prove for all n$$\in$$$$N$$
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$$\in$$$$N$$ 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 $(x+ y)^n$.