Regarding the proof of $J(\chi,\chi^{-1}) = -\chi(-1)$ 
I am having trouble to understand the proof of the following identity for Jacobi Sums:
$$J(\chi,\chi^{-1}) = -\chi(-1)$$
, where $\chi$ is a non-trivial character over $\mathbb{F}_p$ (where $p$ is a prime).

The proof in Ireland-Rosen's book goes as follows:
$$J(\chi,\chi^{-1}) = \sum_{a+b =1 } \chi(a)\chi^{-1}(b) = \sum_{a+b =1 } \chi \biggl(\frac{a}{b}\biggr) = \sum_{a \ne 1} \chi \biggl(\frac{a}{1-a}\biggr) = \sum_{c \ne -1} \chi(c) = -\chi(-1)$$
I do not understand the penultimate equation at all. The book says that it is just a substitution of $c = \frac{a}{1-a}$ and as the $a$ run over all values in $\mathbb{F}_p \setminus \{1\}$ so must the $c$ run over $\mathbb{F}_p \setminus \{-1\}$. I do not understand this. Could you please explain this to me?

Remark: This question is related to my former question. There I learned that
$$\Bigl\{ \frac{a}{a-1} \mid a \ne 1,  a \in \mathbb{F}_p \Bigr\} \ne \Bigl\{ b \mid b \ne -1, b \in \mathbb{F}_p \Bigr\}$$
, which adds to my confusion.


Note that $\frac a{1-a} = -1 + \frac1{1-a}$. As $a$ runs over $\Bbb F_p\setminus\{1\}$, the quantity $1-a$ runs over $\Bbb F_p\setminus\{0\}$, and so $\frac1{1-a}$ also runs over $\Bbb F_p\setminus\{0\}$, and therefore $\frac a{1-a} = -1 + \frac1{1-a}$ runs over $\Bbb F_p\setminus\{-1\}$. That is the reason for the penultimate equality.
(Your former question dealt with the expression $\frac a{a-1}$ rather than $\frac a{1-a}$, and thus is not directly relevant.)
