# Prove that $\sum_{k=1}^{n}\binom{n}{k}=\sum_{k=0}^{n-1}2^{k}$

I know it has something to do with the binomial theorem but I'm not sure how the changed summation limits affect it.

$$\sum_{k=1}^{n}\binom{n}{k}=\sum_{k=0}^{n-1}2^{k}$$

$$\sum_{k=1}^n{n\choose k}=-1+\sum_{k=0}^n{n\choose k}=-1+2^n=\frac{1-2^n}{1-2}=\sum_{k=0}^{n-1}2^k.$$
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k = 1 + \sum_{k=1}^n \binom{n}{k}x^k \implies (1+x)^n\bigg|_{x=1} = \cdots$$
$$\sum_{k=0}^{n-1} 2^k = \frac{2^n - 1}{2-1}$$
Binomial Theorem $$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}.$$ Set $a=b=1$ and get $$2^n=(1+1)^n=\sum_{k=0}^n\binom{n}{k}.$$
$$\sum_{k=1}^{n}\binom{n}{k}=-1+\binom{n}{0}+\sum_{k=1}^{n-1}2^{k}=-1+\sum_{k=0}^{n-1}2^{k}=-1+2^n=$$ $$=2^n-1=\frac{2^n-1}{2-1}=1+2+...+2^{n-1}=\sum_{k=0}^{n-1}2^{k}$$