Proving $\sum\limits_{k=0}^n \binom{n}{k} = 2^n$ by counting in two ways

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Prove $$\sum_{k=0}^n {n \choose k}=2^n$$ by proving that both sides of the formula represent the number of subsets of a set of $n$ elements. For the left side use the addition rule for counting after partitioning the collection of all subsets according to size. And for the right side use the product rule for counting after identifying a subset $A \subset \{1, 2, ..., n\}$ with the sequence of zeros and ones which is the indicator of $A$.

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Hint

As a well-known fact, the number of all subsets of a set $A$ with the cardinality $n$ is $2^n$ since each member of $A$ may or may not be in that subset.

As another insight, any subset of $A$ with $k$ members, must be made up of exactly $k$ distinct members of $A$. How then, can you finish the proof?