. What is the value of the sum of binomial coefficients

$$
\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n-1}+\binom{n}{n} ?
$$
Two ways to do it:

1. We count the subsets of a set with $n$ elements. For each element, there are two possibilities: we take it, or we do not take it. Therefore, the sum in question is $2^{n}$.
2. Another method (suggested by a student): We add $1^{k} \times 1^{n-k}$ (which were of course already there, but invisible!) and use the binomial theorem:

$$
\binom{n}{0} 1^{0} 1^{n}+\binom{n}{1} 1^{1} 1^{n-1}+\binom{n}{2} 1^{2} 1^{n-2}+\cdots+\binom{n}{n-1} 1^{n-1} 1^{1}+\binom{n}{n} 1^{n} 1^{0}=(1+1)^{n}=2^{n}
$$