Sum of the binomial coefficients

Substitute $a=1,\ b=1$ into the binomial theorem $(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$: the left side becomes $(1+1)^n=2^n$, and the right side becomes $\sum_{k=0}^n\binom{n}{k}1^{n-k}1^k=\sum_{k=0}^n\binom{n}{k}$, since $1$ raised to any power is $1$.

Equating the two sides gives exactly $2^n=\sum_{k=0}^n\binom{n}{k}$.

This identity also has a direct combinatorial meaning: $\binom{n}{k}$ counts the $k$-element subsets of an $n$-element set $S$, so $\sum_{k=0}^n\binom{n}{k}$ counts every subset of $S$ of any size at all, i.e. the whole power set, which has exactly $2^n$ elements because each of the $n$ elements is independently either in or out of a subset.