Sum of binomial coefficients

Statement

Let $n$ be a natural number. Then $$\sum_{k=0}^{n} \binom{n}{k} = 2^n.$$

Remarks

For $n = 4$: $1 + 4 + 6 + 4 + 1 = 16$. The identity is the binomial theorem with $x = y = 1$. Combinatorially, both sides count the subsets of an $n$-element set. The number of subsets of a finite set is $2^n$, and $\binom{n}{k}$ of them have size $k$.

Explanations

L1 human

The numbers in row $n$ of Pascal's triangle add up to $2^n$. This is because each subset of an $n$-element set has some size $k$ from $0$ to $n$, and there are $2^n$ subsets in total.