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 $n = 3, 1 + 2 + 4 = 7 = 8 - 1$.

$1 + 2+ 4 + 8  + ......... + 2^{n-2} + 2^{n-1}=$
$1 + 1 + 2+ 4 + 8  + ......... + 2^{n-2} + 2^{n-1} - 1=$
$2 + 2+ 4 + 8  + ......... + 2^{n-2} + 2^{n-1} - 1=$
$4+ 4 + 8  + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$8 + 8  + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$16  + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
.......
$2^j + 2^j + 2^{j+1}+ ...... + 2^{n-2} + 2^{n-1} - 1=$
$2^{j+1} + 2^{j+1}+.....+ 2^{n-2} + 2^{n-1} - 1=$
$2^{j+2} + ..... + 2^{n-2} + 2^{n-1} - 1=$
..........
$2^{n-2} + 2^{n-2} + 2^{n-1} - 1=$
$2^{n-1} + 2^{n-1} - 1 = $
$2^n - 1$.
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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 
$2^{n-1} + 2^{n-2} + 2^{n-3}+........ + 16+ 8 + 4 + 2 + 1 = 2^n - 1$
