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.

There are a few points where I'd suspect someone could take issue with your answer:
The Induction Hypothesis is supposed to be the idea that given the hypothesis holds for a set of values from the base case up to an n that you could use that to prove the n+1 case.  You seem to imply that the Induction Hypothesis holds for all natural numbers which is a bit of an overstatement to my mind.
Lastly, where is the use of the induction hypothesis in the last step?  Where do you use the $2^n$-1 part that should be part of the solution.

$\sum_{k=1}^{n+1} \binom {n+1} {k} = \sum_{k=1}^{n} \binom {n+1} {k} + \binom{n+1}{n+1} = (n+1)\sum_{k=1}^{n}\binom {n} {k} + 1 = (n+1)(2^n-1)+1 = n2^n-n+2^n-1+1 = 2^n(n+1)-n$ which still has to be simplified into the final result but this at least is using the hypothesis.
