derivation of Gauss sum up to a sign

The Gauss sum can be easily evaluated up to a sign by squaring the original series

$g_1^2(\chi )$$={\displaystyle \sum _{s\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{s}{p}}\right)e^{2\pi is/p}{\displaystyle \sum _{t\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{t}{p}}\right)e^{2\pi it/p}$

$={\displaystyle \sum _{s,t\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{st}{p}}\right)e^{2\pi i(s+t)/p}$

and summing over a new variable $n=s^{-1}t(modp)$

$={\displaystyle \sum _{s,n\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{n}{p}}\right)e^{2\pi i(s+ns)}$

$={\displaystyle \sum _{n\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{n}{p}}\right){\displaystyle \sum _{s\in \mathbb{Z}/p\mathbb{Z}}}{e}^{2\pi is(n+1)}$

$={\displaystyle \sum _{n\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{n}{p}}\right)(q[n\equiv -1(modp)]-1)$

$=p\left({\displaystyle \frac{-1}{p}}\right)-{\displaystyle \sum _{n\in \mathbb{Z}/p\mathbb{Z}}}\left({\displaystyle \frac{n}{p}}\right)$