Let $p$ be an odd prime and $a_1, a_2, a_3, \ldots , a_n$ be integers such that $p$ does not divide any of these integers. Prove that
$$ (\frac{a_1}{p}) \times (\frac{a_2}{p}) \times \ldots \times (\frac{a_n}{p}) = (\frac{a_1 \times a_2 \times \ldots \times a_n}{p}) $$
where $(\frac{a}{p})$ is the Legendre symbol.
We will do this by indiction. The statement $P(n)$ is
$$ (\frac{a_1}{p}) \times (\frac{a_2}{p}) \times \ldots \times (\frac{a_n}{p}) = (\frac{a_1 \times a_2 \times \ldots \times a_n}{p}) $$
We need to show the base case is true $P(1)$, and the inductive step $P(n) \implies P(n+1)$.
Base Case $P(1)$
The base case $P(1)$ is trivially true
$$ (\frac{a_1}{p}) = (\frac{a_1}{p}) $$
Inductive Step $P(n) \implies P(n+1)$
We assume the induction hypothesis $P(n)$ is true.
We proceed as follows
$$ \begin{align} (\frac{a_1}{p}) \times (\frac{a_2}{p}) \times \ldots \times (\frac{a_n}{p}) \times (\frac{a_{n+1}}{p}) & = \overbrace{\ (\frac{a_1 \times a_2 \times \ldots \times a_n}{p})}^{\text{induction hypothesis}} \times (\frac{a_{n+1}}{p}) \\ \\ & = (\frac{a_1 \times a_2 \times \ldots \times a_n \times a_{n+1}}{p}) \end{align}$$
The last step uses Proposition (7.9)(c), and uses the fact that $p \not \mid (a_1 \times a_2 \times \ldots \times a_n)$ and $p \not \mid a_{n+1}$.
And so $P(n) \implies P(n+1)$.
We have shown by induction that
$$ (\frac{a_1}{p}) \times (\frac{a_2}{p}) \times \ldots \times (\frac{a_n}{p}) = (\frac{a_1 \times a_2 \times \ldots \times a_n}{p}) $$
for natural number $n$, odd prime $p$ which does not divide any of the $a_i$.