Induction proof of $\sum_{k=1}^{n} \binom n k = 2^n -1 $ 
Prove by induction: $$\sum_{k=1}^{n} \binom n k = 2^n -1 $$ for all $n\in \mathbb{N}$.

Today I wrote calculus exam, I had this problem given.
I have the feeling that I will get $0$ points for my solution, because I did this: 
Base Case: $n=1$
$$\sum_{k=1}^{1} \binom 1 1 = 1 =  2^1 -1 .$$
Induction Hypothesis: for all $n \in \mathbb{N}$:
$$\sum_{k=1}^{n} \binom n k = 2^n -1 $$
Induction Step: $n \rightarrow n+1$
$$\sum_{k=1}^{n+1} \binom {n+1} {k} = \sum_{k=1}^{n} \binom {n+1} {k} + \binom{n+1}{n+1} = 2^{n+1} -1$$
Please show me my mistake because next time is my last chance in this class.

suppose that $$\sum_{k=1}^{n} \binom n k = 2^n -1 $$
then
$$\sum_{k=1}^{n+1} \binom {n+1}{k} =\sum_{k=1}^{n+1}\Bigg( \binom {n}{ k} +\binom{n}{k-1}\Bigg)=$$
$$=\sum_{k=1}^{n+1} \binom {n}{ k} +\sum_{k=1}^{n+1}\binom{n}{k-1}=$$
$$=\sum_{k=1}^{n} \binom {n}{ k}+\binom{n}{n+1}+\sum_{k=0}^{n} \binom {n}{k}=$$
$$=\sum_{k=1}^{n} \binom {n}{ k}+\binom{n}{n+1}+\binom{n}{0}+\sum_{k=1}^{n} \binom {n}{k}=$$
$$=2\sum_{k=1}^{n} \binom {n}{ k}+\binom{n}{n+1}+\binom{n}{0}=$$
since $\binom{n}{n+1}=0, \binom{n}{0}=1$
$$=2\sum_{k=1}^{n} \binom {n}{ k}+1=2(2^n-1)+1=2^{n+1}-2+1=2^{n+1}-1$$
