Proving $\sum_{i=0}^n \binom{n}{i} = 2^n$ by math induction I am having some trouble using math induction to prove the following problem:

$$\sum_{i=0}^n \binom{n}{i} = 2^n$$ Where n $\geq$ 0

I know the first thing with math induction is substitute the base value for n. 

Which we get $$\sum_{i=0}^0 \binom{0}{0} = 2^0$$ which equals to 1 on the left hand side and 1 on the right hand side. The base case is now verified. We move on to substituting n for k (assuming k is true) like so:

$$\sum_{i=0}^k \binom{k}{i} = 2^k$$
If k is true, k + 1 should be true as well. So we must prove k + 1 now:
$$\sum_{i=0}^{k+1} \binom{k+1}{i} = 2^{k+1}$$

This is the part I am having trouble on. Are all my steps correct so far? If so, how do I finish proving this problem?
You can take advantage of the recurrence relation:

$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k},1 \leq k \leq n-1$$

$${n \choose n} = {n \choose 0} = 1, n \geq 0.$$

Then split apart the $k+1$ sum and watch the indices to make sure you include all of the terms:

$$\sum_{j=0}^{k+1} {k+1 \choose j} \\ = \sum_{j=0}^{k} {k \choose j-1} + \sum_{j=0}^{k} {k \choose j} + {k+1 \choose k+1}  \\ = \left[\left(\sum_{j=0}^{k} {k \choose j}\right) - {k \choose k} \right] + \sum_{j=0}^{k} {k \choose j} + {k+1 \choose k+1} \\ =2^k - 1 + 2^k + 1 \\ = 2 \cdot 2^k \\ = 2^{k+1}.$$
If

$$1+4+6+4+1=2^4,$$

then by the well-known additive property of the binomial coefficients,

$$\begin{align}2^4+2^4=\\1+&amp;4\ +6\ +4\ +1\ +\\
&amp;1\ +4\ +6\ +4\ +1=\\
1+&amp;5+10+10+5+1=2^5.\end{align}$$
This should be accepted as a proof by induction, as it unambiguously shows the pattern, which can be generalized for other $n$. (The first identity can also be used for the base case, but $1+1=2^1$ or $1=2^0$ are even better.)