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. 
 A: Initial comment: First of all, +1 for effort. Your questions almost always show a lot of it. Also, I can see you are trying to actively improve based on questions like this, where you are clearly trying to implement the advice given by users on here. That's great. Based on several of your questions, your induction ones anyway, it seems that one of the chief struggles you have is in actually writing the proof well (that is, making it clear, polished, etc.). Thus, at the end of this post, I'll provide you with a template for writing induction proofs adapted from David Gunderson's Handbook of Mathematical Induction (something tells me you would really like this book by the way). It helped me immensely when I was starting out. 

General case of your problem: Fix $r\in\mathbb{R}, r\neq 1$, and for $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : 1+r+r^2+\cdots+r^n=\frac{r^{n+1}-1}{r-1}.\tag{1}
$$
Your specific problem is solved by setting $r=2$ in $(1)$. I actually proved $(1)$ by induction some time ago, but I cannot seem to find that question (either it was deleted or I simply cannot find it). Regardless, you should at least know that your problem is really an instance of a more general scenario. 

Your specific problem: I am going to provide what I think is a nice write-up for your exact problem (note: only read on if you want the full solution). My main reason for providing it is to show you how a well-written proof would look (IMO anyway). 
For $n\geq 0$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^n 2^i = 2^{n+1}-1.
$$
Base step ($n=0$): $S(0)$ says that $2^0 = 2^1-1$, which is true.
Induction step ($S(k)\to S(k+1)$): Fix some $k\geq 0$ and suppose that
$$
S(k) : \sum_{i=1}^k 2^i = 2^{k+1}-1
$$
is true. To be proved is that
$$
S(k+1) : \sum_{i=1}^{k+1} 2^i = 2^{k+2}-1
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1} 2^i &= \sum_{i=1}^k 2^i+2^{k+1}\tag{by definition of $\Sigma$}\\[0.75em]
&= (2^{k+1}-1)+2^{k+1}\tag{by $S(k)$}\\[0.75em]
&= 2\cdot 2^{k+1} - 1\tag{group like terms}\\[0.75em]
&= 2^{k+2}-1,
\end{align}
we see that the right-hand side of $S(k+1)$ follows. This completes the inductive step.
Thus, by mathematical induction, for all $n\geq 0, S(n)$ holds. $\Box$

Template: As promised, here's a template you can use from now on that should greatly assist you in writing up clear induction proofs. 

A: 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.
