Proof that $\sum\limits_{k = 0}^n {n\choose k} =2^n$ using Binomial Expansion Formula

Question

HW problem here. Not sure how to even start on it.

Prove that $$\sum\limits_{k = 0}^n {n\choose k} =2^n$$

Any help is appreciated.

For Search purposes:

(Hint: Use the binomial expansion mentioned on p. 87.)
An Introduction to Mathematical Statistics and It's Applications 2.6.58

Answer 1

Hint: it should be mentioned, on p. 87, that $$\forall x,y\ \ \sum_{k=0}^n\binom nk x^ky^{n-k} = (x+y)^n, $$

Answer 2

Second hint

Use the binomial expansion in $$(1+1)^n$$

Answer 3

If your aim is to use the binomial theorem, the other answers will do, but a nice way to see this is to count the number of subsets of a set containing $n$ elements. To count this, we can run through each element and choose to include or exclude it from the subset. This is two choices for each element, giving $2^n$ subsets. But we can also count this by adding up the number of subsets of size $k$ for each $k$ ranging from $0$ to $n$; this is $\sum {n \choose k}$.