$2^n-1 = \sum_{jI am having trouble understanding the following analysis after we arrived to the conclusion:
$2^k - \sum_{j=0}^{j=k-1}2^j = 1$
after arriving to the conclusion, they say, I think to explain that the left side is equal to 1: 
"In fact, by induction on $n$ in $2^n-1 = \sum_{j<n}2^j$, in fact, $2^{n+1} - 1 = 2(2^n - 1) + 1 = 1 + \sum_{0<j<n+1}2^j = \sum_{j<n+1}2^j.$"
I am not even sure if the wrote the last part to prove 
$2^k - \sum_{j=0}^{j=k-1}2^j = 1$, is it to do that? If it is so, why?
Also, I do not see how we pass from $2(2^n - 1) + 1$ to $1 + \sum_{0<j<n+1}2^j$, and from $1 + \sum_{0<j<n+1}2^j$ to $\sum_{j<n+1}2^j$. 
Thank you!
 A: Here is my solution by using induction explanation:


*

*For $n=1$, we have $\displaystyle 2^{1}-1=1=\sum_{i=0}^{n-1}{2^{i}}$. So when $n=1$, the equation holds.

*We suppose the equation holds when $n=k$, that is to say $\displaystyle 2^{k}-1=\sum_{i=0}^{k-1}{2^{i}}.$
For $n=k+1$, we have $\displaystyle 2^{k+1}-1=2^{k}+2^{k}-1=2^{k}+\sum_{i=0}^{k-1}{2^{i}}=\sum_{i=0}^{k}{2^{i}}.$
So we can get from 1. and 2. that the statement is true.
Actually, we needn't prove it in a induction explanation way, we can simply add all the number up in the right side by using subtract dislocation method and we can get the answer.
A: Here's a less apocalyptic induction solution.
Let's note for all $n \in \mathbb{N}$, $P(n) : 2^n-1 = \sum_{j<n}2^j$.
$\sum_{j<0}2^j = 0 = 2^0-1$ so $P(0)$ is true.
Let's suppose that $\exists n \in \mathbb{N}$ such as $P(n)$ is true.
So we have $2^n-1 = \sum_{j<n}2^j$.
$\sum_{j<n+1}2^j = 2^n+\sum_{j<n}2^j = 2^n +2^n-1$ according to the recurrence hypothesis.
So $\sum_{j<n+1}2^j = 2^{n+1}-1$ and consequently, $P(n+1)$ is true.
By recurrence over $\mathbb{N}$, $\forall n \in \mathbb{N}$, $P(n)$ is true.
Note : There is a much more straightforward proof : if we define $S_n$ to be $\sum_{j<n}2^j$, $2S_n = 2\sum_{j<n}2^j = \sum_{j<n}2^{j+1} = \sum_{j\in [1,n]}2^j = -1+2^{n}+\sum_{j<{n}}2^j = -1+2^{n}+S_n$.
So $S_n = 2^n-1$.
