Proving $\sum_{i=0}^n 2^i=2^{n+1}-1$ by induction. Firstly, this is a homework problem so please do not just give an answer away. Hints and suggestions are really all I'm looking for.
I must prove the following using mathematical induction:

For all $n\in\mathbb{Z^+}$, $1+2+2^2+2^3+\cdots+2^n=2^{n+1}-1$.

This is what I have in my proof so far:

Proof: Let $p\left(n\right)$ be $\sum_{j=0}^n 2^j=1+2+2^2+\cdots+2^n=2^{n+1}-1$. Then $P\left(0\right)$ is 
  \begin{align} \sum\limits_{j=0}^0 2^j=1=2^1-1.\tag{1} \end{align} Hence
   $P\left(0\right)$ is true. Next, for some $k\in\mathbb{Z^+}$, assume
   $P\left(k\right)$ is true. Then \begin{align}
 \sum\limits_{j=0}^{k+1}2^j&=\left(\sum\limits_{j=0}^k
 2^j\right)+2^{k+1}\tag{2}\\ &=2^{k+1}-1+2^{k+1}\tag{3} \\
&=2\cdot 2^{k+1}-1\tag{4}\\
&=2^{k+2}-1\tag{5}\\
&=2^{\left(k+1\right)+1}-1.\tag{6}
\end{align}
  Therefore, $P\left(n\right)$ holds for all $n$.$\;\;\;\;\blacksquare$

Is everything right? My professor went over the methodology in class quite fast (and I had a bit too much coffee that night) and I want to make sure I have it down for tomorrow night's exam. 

Your problem has already been answered in details. I would like to point out that if you want to train your inductive skills, once you have a solution to your problem, you still can explore it from many other sides, especially using visual proofs that can be quite effective. And find other proofs. Some are illustrated in An Invitation to Proofs Without Words. I will try to describe two of them related to your case.
First, $2^k$ can be represented in binary with only zeros, except for a $1$ in the $(k+1)^{\textrm{th}}$ position from the right:
 - 01: 00000001
 - 02: 00000010
 - 04: 00000100
 - 08: 00001000
 - 16: 00010000

So their sum will add the columns. Since there is exactly one $1$, you do not have carries, and hence get ones up to the $(K+1)^{\textrm{th}}$ position, and the number you are looking for is:
 - ??: 00011111

The induction may appear as: if you have a binary digit with ones up to the $(K+1)^{\textrm{th}}$ position, adding one to it yields a number with a single one in the $(K+2)^{\textrm{th}}$ position, for instance:
 - 00011111+0000001=0010000 

So your mysterious number is $2^{K+2}-1$.
The second proof is also illustrated with fractions of powers of two (see the top-right picture in 1/2 + 1/4 + 1/8 + 1/16 + ⋯). One can observe that powers of two are either:


*

*pure squares with sides $2^k\times 2^k$

*double-square rectangles with  sides $2^k\times 2^{k+1}$


Here, you can get a double-induction, that often happens in series: one property
for odd indices, another property for even indices. The two properties are: 


*

*the sum of powers of two up to a square ($k=2K$) give the next rectangle minus one,

*the sum of powers of two up to a rectangle ($k=2K+1$) give the next square minus one,


which you can prove by alternated inductions.
