Proof: Still by (30), we have
$$\begin{aligned} \left(\frac{a}{p}\right) \equiv & a^{\frac{p-1}{2}}=a_{1} \frac{p-1}{2} a_{2} \frac{p-1}{2} \cdots a_{n} \frac{p-1}{2} \equiv\left(\frac{a_{1}}{p}\right)\left(\frac{a_{1}}{p}\right) \\ & \cdots\left(\frac{a_{n}}{p}\right)(\bmod p) \end{aligned}$$

Thus, we get $\left(\frac{a}{p}\right)-\left(\frac{a_{1}}{p}\right)\left(\frac{a_{2}}{p}\right) \ldots\left(\frac{a_{n}}{p}\right) \equiv 0(\bmod p)$, but it is easy to see that
$$\left|\left(\frac{a}{p}\right)-\left(\frac{a_{1}}{p}\right)\left(\frac{a_{2}}{p}\right) \ldots\left(\frac{a_{n}}{p}\right)\right| \leqslant 2,$$

and $p \geqslant 3$, so it must be that
$$\left(\frac{a}{p}\right)=\left(\frac{a_{1}}{p}\right)\left(\frac{a_{2}}{p}\right) \ldots\left(\frac{a_{p}}{p}\right),$$

which is the result we wanted to prove.