Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to ${2^n}{-1}$. Please explain in quotations!
"Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to $2^n-1$." In a very simple case, for $n = 3, 1 + 2 + 4 = 7 = 8 - 1$.
 A: They mean to the formula $2^0+2^1+2^2+\ldots+2^{n-1}=2^n-1$ which is a particular case of 
$$
q^0+q^1+q^2+\ldots+q^{n-1}=\frac{q^n-1}{q-1}\quad(q\neq1)
$$
for $q=2$. 
A: HINT:
Suppose $$\sum_{i=0}^{n-1} 2^i=2^n-1$$
is true, then $$\sum_{i=0}^{n}2^i=\sum_{i=0}^{n-1}2^i+2^{n}=2^{n}-1+2^{n}=2\times2^n-1=2^{n+1}-1$$
A: Since we don't know what $1+2+4+\cdots+2^{n-1}$ is yet, we'll call that number $S$. So we have $S=1+\cdots+2^{n-1}$. Multiplying by $2$ on both sides, we get
$$
2S=2+4+8+\cdots+2^{n-1}+2^n
$$
Here comes the trick: calculate $2S-S$:
$$
\begin{align}
2S-S=&2+4+8+\cdots+2^{n-1}+2^n\\-(1+&2+4+8+\cdots+2^{n-1})
\end{align}
$$
and we see that all but two of the terms disappear, and we're left with $2S-S=2^n-1$.
A: If you’re familiar with binary, simply note that
$$2^n-1 = {1\underbrace{000\dots00}_{\text{$n$ zeros}}}\text{$_2$} - 1 = \underbrace{111\dots11}_{\text{$n$ ones}}\text{$_2$}$$
A: $1 + 2+ 4 + 8  + ......... + 2^{n-2} + 2^{n-1}=$
$1 + 1 + 2+ 4 + 8  + ......... + 2^{n-2} + 2^{n-1} - 1=$
$2 + 2+ 4 + 8  + ......... + 2^{n-2} + 2^{n-1} - 1=$
$4+ 4 + 8  + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$8 + 8  + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$16  + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
.......
$2^j + 2^j + 2^{j+1}+ ...... + 2^{n-2} + 2^{n-1} - 1=$
$2^{j+1} + 2^{j+1}+.....+ 2^{n-2} + 2^{n-1} - 1=$
$2^{j+2} + ..... + 2^{n-2} + 2^{n-1} - 1=$
..........
$2^{n-2} + 2^{n-2} + 2^{n-1} - 1=$
$2^{n-1} + 2^{n-1} - 1 = $
$2^n - 1$.
========
Or another way:
$2^n = 2*2^{n-1} = 2^{n-1} + 2^{n-1}=$
$2^{n-1} + 2*2^{n-2} = 2^{n-1} + 2^{n-2} + 2^{n-2}=$
$2^{n-1} + 2^{n-2} + 2*2^{n-3} = 2^{n-1} + 2^{n-2} + 2^{n-3}+2^{n-3}=$
........
$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 8=$
$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 4=$
$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 2=$
$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 1+1=$
$2^n$
So 
$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 1 = 2^n - 1$
