TITLE: Proving by induction that $ \sum_{k=0}^n{n \choose k} = 2^n$
QUESTION [24 upvotes]: Prove by induction that for all $n \ge 0$:
$${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$
In the inductive step, use Pascal’s identity, which is:
$${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$
I can only prove it using the binomial theorem, not induction.

REPLY [2 votes]: Since someone decided to revive this 6 year old question, you can also prove this using combinatorics
Take n elements and count how many ways there are to put these two elements into 2 different containers (A and B)
a) Every element can take two states: it's either in A or in B. By one of the first theorems in combinatorics (the one about separating a task into steps), this gives n steps with 2 options at each step, so we get $2^n$
b) We can also think of adding up all the ways in which we can have k elements in A and n-k elements in B, for $0 \le k \le n$. For each k, this number is: "choose k elements that will go into A". Then the other n-k elements automatically go to B. 
So for each k this number is just ${n \choose k}$. Now, we add this for all k reviously mentioned to get $\sum_{k = 0}^n {n \choose k}$
And thus, $2^n = \sum_{k = 0}^n {n \choose k}$