Why does $\sum_{k=1}^{n-1}{n \choose k} = -1 -1 + \sum_{k=0}^{n} {n \choose k}$ by the binomial theorem?

Question

Why does $$\sum_{k=1}^{n-1}{n \choose k} = -1 -1 + \sum_{k=0}^{n} {n \choose k}$$ by the binomial theorem?

The binomial theorem is $(x+y)^n = \sum_{k=0}^{n} {n \choose k} x^ky^{n-k}$

Answer 1

This has not much to do with the binomial theorem, but follows directly from $\binom{n}{0} = \binom{n}{n} = 1$.

One can use the binomial theorem to prove somewhat more interestering identities involving binomial coefficients, such as $$ \sum_{k=0}^n \binom{n}{k} = 2^n \qquad \text{and} \qquad \sum_{k=0}^n (-1)^k \binom{n}{k} = 0,$$ by plugging in $x=y=1$ and $x=-y=1$, respectively.