Why it's $2^8-127$, not $2^8-(-127)$ when "two's complement of $a$" is $2^n-a$ in binary notation? "Table 2.5.1 Powers of 2
$$\begin{array} {|c|c|c|c||c|}
\hline
\text {Power of 2} & 2^{10}& 2^9 & 2^8 & 2^7& 2^6&2^5 &2^4 &2^3  &2^2 &2^1 &2^0 \\ \hline
2 & 1024 & 512 & 256 & 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1\\ \hline
\end{array} $$"
"Definition Given a positive integer a, the two's complement of a relative to a fixed bit length n is the n-bit binary representation of $2^n-a$"
p. 84 
I don't understand the last five rows in Table 2.5.2, the above definition says the two's complement of a positive integer, but the below shows the complement of a negative integer, and I don't understand why negative integer $a$ becomes $-a$ unlike the definition $2^n-a$, for example one of the five rows, why it's $2^8-127$, not $2^8-(-127)$?  
"Table 2.5.2 $$\begin{array} {|c|c|c|c||c|}
\hline
Integer & \begin{array}{c}\text {8-Bit Representation}\\\text{(ordinary 8-bit binary}\\\text{ if nonnegative or 8-bit two's}\\\text{complement of absolute value if negative)}\end{array} & \begin{array}{c}\text {Decimal form of}\\\text{Two's complement}\\\text{for Negative Integers)}\end{array} \\ \hline
127 & 01111111 &   \\
126 & 01111110 &   \\
. & . &   \\
. & . &   \\
. & . &   \\
2 & 00000010 &   \\
1 & 00000001 &   \\
0 & 00000000 &    \\
-1 & 11111111 & 2^8-1   \\
-2 & 11111110 & 2^8-2  \\
. & . &   \\
. & . &   \\
. & . &   \\
-127 & 10000001 & 2^8-127\\
-128 & 10000000 & 2^8-128\\ \hline
\end{array} $$ p.87"
Source: Discrete Mathematics with Applications by Susanna Epp

Taking two's complement implies inversing a number ie binary digits ie if $0$ then $1$ and vice -versa and adding $1$ to LSB so binary $8$ -bit representation of $1$ is $00000001$ so taking inverse we get $11111110$ and adding $1$ we get $11111111$  if on subtraction the answer is negative then its in the form of 2's complement and if extra bit is generated its positive and we discard it 
