How to prove that $\sum\limits_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$ using the Binomial Theorem? I have this proposition:
$$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$
How can I prove that? How to use the Binomial Theorem to solve that?

Use inductive approach
For n = 0: $$\binom{1}{1} = 1 = 2^1 - 1$$ => TRUE
Assume that for n>=0, the following (*) is true $$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$ 
We need prove: $$\sum_{k=0}^{n+1} \binom{n+2}{k+1} = 2^{n+2} - 1$$
Remember: $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$
So: $$\binom{n+2}{k+1} = \binom{n+1}{k+1} + \binom{n+1}{k}$$
=> $$\sum_{k=0}^{n+1} \binom{n+2}{k+1} = \sum_{k=0}^{n+1}\binom{n+1}{k+1}+\sum_{k=0}^{n+1}\binom{n+1}{k}$$
$$=\binom{n+1}{n+2} + \sum_{k=0}^{n}\binom{n+1}{k+1} + \sum_{k=0}^{n}\binom{n+1}{k+1} - \binom{n+1}{0}$$
$$=2*\sum_{k=0}^{n}\binom{n+1}{k+1} - 1 = 2*2^{n+1} - 1 = 2^{n+2}-1$$
That is what to be proven.
