We use the same notation as in the proof of Gauss' Lemma. Consider the set of integers
\begin{equation*}
S = \left\{a, 2a, 3a, ..., \frac{p-1}{2}a\right\}.
\end{equation*}
Divide each of these multiples of \(a\) by \(p,\) to get
\begin{equation*}
ka = \left\lfloor\frac{ka}{p}\right\rfloor p+t_k,
\end{equation*}
where \(0\leq t_k \lt p-1.\) If \(t_k\lt p/2,\) then it is one of the integers \(r_1,..., r_m;\) if \(t_k \gt p/2,\) then it is one of the integers \(s_1,..., s_n.\)
Summing the elements of \(S,\) we have
\begin{equation*}
\sum_{k=1}^{(p-1)/2} ka =\sum_{k=1}^{(p-1)/2}\left\lfloor\frac{ka}{p}\right\rfloor p+t_k =\sum_{k=1}^{(p-1)/2}\left\lfloor\frac{ka}{p}\right\rfloor p+\sum_{k=1}^{m}r_k+\sum_{k=1}^{n}s_k.
\end{equation*}
As in Gauss' Lemma we have that the \((p-1)/2\) numbers
\begin{equation*}
r_1,..., r_m, p-s_1,..., p-s_n
\end{equation*}
are just a rearrangement of \(1, 2,...,(p-1)/2.\) So
\begin{equation*}
\sum_{k=1}^{(p-1)/2} k =\sum_{k=1}^{m}r_k+\sum_{k=1}^{n}(p-s_k)=pn+\sum_{k=1}^{m}r_k-\sum_{k=1}^{n}s_k.
\end{equation*}
Subtracting this equation from the last gives
\begin{equation*}
(a-1)\sum_{k=1}^{(p-1)/2} k =p\left(\sum_{k=1}^{(p-1)/2}\left\lfloor\frac{ka}{p}\right\rfloor -n\right) +2\sum_{k=1}^{n}s_k.
\end{equation*}
Since \(a\) is an odd integer, we have that \(p \equiv a \equiv 1\pmod{2},\) so that modulo \(2\) we have,
\begin{equation*}
(0) \cdot \sum_{k=1}^{(p-1)/2}k \equiv 1\cdot \left(\sum_{k=1}^{(p-1)/2} \left\lfloor\frac{ka}{p}\right\rfloor-n\right)\pmod{2}.
\end{equation*}
This means that the sum and \(n\) have the same parity, so that
\begin{equation*}
(-1)^n=(-1)^{\sum_{k=1}^{(p-1)/2}\lfloor ka/p\rfloor},
\end{equation*}
thus, by Gauss' Lemma
\begin{equation*}
(a/p)=(-1)^{\sum_{k=1}^{(p-1)/2}\lfloor ka/p\rfloor}.
\end{equation*}