Overflow in Adder Circuits
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[Two's complement:] for a full adder of n bits the range is [-2^(n-1)
, 2^(n-1)-1]. What I don't get is why is -2^(n) also in the list.

\$ -2^n \$ isn't in the range you (correctly) quoted. Were you thinking of \$ -2^{n-1} \$ ?
I trust that you are happy that with \$ n \$ bits we can represent \$ 2^n \$ numbers. We can divide that range into 2 equal halves each representing \$ 2^{n-1} \$ numbers.
For negative numbers we can use one half of the range to represent numbers from  \$ -1 \$ to \$ -2^{n-1} \$.
However for non-negative numbers we need to use one of the bit patterns from the other half of the range to represent  \$ 0 \$ so our maximum positive number is reduced by  \$ 1 \$ compared to the magnitude of the most negative number. So the non-negative range runs from \$ 0 \$ to \$ 2^{n-1}-1 \$.
Hence the total range is \$ [ 2^{n-1} ,   2^{n-1}-1 ] \$ as you quoted. In two's complement representation the most negative number will always be even and will have a magnitude (absolute value) one more than the most positive number, which will always be odd (for  \$ n &gt; 1 \$).