Product of Gauss sums $\tau_p,\tau_q$ Let $p,q$ be different odd primes, and let $\tau_p = \sum\limits_{a=1}^{p}\left(\frac{a}{p}\right)e^{\frac{2\pi ia}{p}}$, $\tau_p = \sum\limits_{b=1}^{q}\left(\frac{b}{q}\right)e^{\frac{2\pi ib}{q}}$. I need to prove that
$$
(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\tau_p\tau_q = \sum\limits_{c=1}^{pq}\left(\frac{c}{p}\right)\left(\frac{c}{q}\right)e^{\frac{2\pi ic}{pq}}
$$
So, we have that
$$
\tau_p\tau_q = \sum\limits_{a=1}^{p}\sum\limits_{b=1}^q\left(\frac{a}{p}\right)\left(\frac{b}{q}\right)e^{\frac{2\pi i(aq+bp)}{pq}}
$$
If we use quadratic reciprocity, we get that
$$
(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\tau_p\tau_q = \sum\limits_{a=1}^p\sum\limits_{b=1}^q\left(\frac{aq}{p}\right)\left(\frac{bp}{q}\right)e^{\frac{2\pi i(aq+bp)}{pq}}
$$
This immediately implies that a definition $c = aq+bp$ is required. But, I can't seem to work out the index swap that will allow me to change from a double summation to a single summation. I've tried to prove that every $c\mod pq$ has a single decomposition of $aq + bp\mod pq$, where $1\leq a\leq p, 1\leq b\leq q$. But I was only able to prove uniqueness (using the Chinese remainder theorem). Also, I don't see how this helps me.

By the Chinese Remainder Theorem the quantity $c=aq+bp$ ranges over all the residue classes modulo $pq$, when $a$ ranges over residue classes modulo $p$ and $b$ over residule classes modulo $q$. Also observe that
$$
\left(\frac c p\right)=\left(\frac {aq+bp}p\right)=\left(\frac{aq}p\right)
$$
because $c\equiv aq\pmod p$. Similarly 
$$
\left(\frac c q\right)=\left(\frac {bp}q\right).
$$
Therefore
$$
\sum\limits_{a=1}^p\sum\limits_{b=1}^q\left(\frac{aq}{p}\right)\left(\frac{bp}{q}\right)e^{\frac{2\pi i(aq+bp)}{pq}}
=\sum_{c=1}^{pq}\left(\frac cp\right)\left(\frac cq\right)e^{\frac{2\pi i c}{pq}}.
$$
