Let $p$ be an odd prime. Prove that
$$ \left (1 × 3 × 5 × ⋯ × (p− 2) \right )^2 \equiv (−1)^{\frac{p+1}{2}} \pmod p$$
There are two cases to consider, an even number of multiplicands, and an odd number of multiplicands.
Case: Even number of multiplicands.
In this case the sequence of multiplicands is
$$1,\ldots, \frac{p-3}{2}, \frac{p+1}{2}, \ldots, (p-2)$$
And so
$$ \begin{align} \biggl(1 \times 3 \times \ldots \times (p− 2) \biggr )^2 & \equiv \biggl (1 \times 3 \times \ldots (\frac{p-7}{2}) \times (\frac{p-3}{2}) \times (\frac{p+1}{2}) \times (\frac{p+5}{2}) \ldots \times (p-4) \times (p− 2) \biggr )^2 \pmod p \\ \\ & \equiv \biggl (1 \times 3 \times \ldots (\frac{p-7}{2}) \times (\frac{p-3}{2}) \times (\frac{-p+1}{2}) \times (\frac{-p+5}{2}) \ldots \times (-4) \times (− 2) \biggr )^2 \pmod p \\ \\ & \equiv \biggl (1 \times 3 \times \ldots (\frac{p-7}{2}) \times (\frac{p-3}{2}) \times (\frac{p-1}{2}) \times (\frac{p-5}{2}) \ldots \times (+4) \times (+2) \biggr )^2 \pmod p \\ \\ & \equiv \biggl( (\frac{p-1}{2})! \biggr )^2 \pmod p \end{align} $$
Case: Odd number of multiplicands.
In this case the sequence of multiplicands is
$$1,\ldots, \frac{p-5}{2}, \frac{p-1}{2}, \frac{p+3}{2}, \ldots, (p-2)$$
And so
$$ \begin{align} \biggl(1 \times 3 \times \ldots \times (p− 2) \biggr )^2 & \equiv \biggl (1 \times 3 \times \ldots (\frac{p-5}{2}) \times (\frac{p-1}{2}) \times (\frac{p+3}{2}) \ldots \times (p-4) \times (p− 2) \biggr )^2 \pmod p \\ \\ & \equiv \biggl (1 \times 3 \times \ldots (\frac{p-5}{2}) \times (\frac{p-1}{2}) \times (\frac{-p+3}{2}) \ldots \times (-4) \times (− 2) \biggr )^2 \pmod p \\ \\ & \equiv \biggl (1 \times 3 \times \ldots (\frac{p-5}{2}) \times (\frac{p-1}{2}) \times (\frac{p-3}{2}) \ldots \times (+4) \times (+2) \biggr )^2 \pmod p \\ \\ & \equiv \biggl ( (\frac{p-1}{2})! \biggr)^2 \pmod p \end{align} $$
So in both cases we have
$$ \biggl(1 \times 3 \times \ldots \times (p− 2) \biggr )^2 \equiv \biggl ( (\frac{p-1}{2})! \biggr)^2 \pmod p $$
From exercise (4.2).16 we have
$$ -1 \equiv (-1)^{\frac{p-1}{2}} \biggl ( (\frac{p-1}{2})! \biggr)^2 \pmod p $$
Multiplying both sides by $(-1)^{\frac{p-1}{2}}$
$$ (-1)^{\frac{2}{2}} \times (-1)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} (-1)^{\frac{p-1}{2}} \biggl ( (\frac{p-1}{2})! \biggr)^2 \pmod p $$
Noting that $p-1$ is even, and so $(-1)^{\frac{p-1}{2}} (-1)^{\frac{p-1}{2}} \equiv (-1)^{p-1} \equiv 1 \pmod p$.This gives us
$$ (-1)^{\frac{p+1}{2}} \equiv \biggl ( (\frac{p-1}{2})! \biggr)^2 \pmod p $$
Which leads immediately to our desired conclusion
$$ \left (1 × 3 × 5 × ⋯ × (p− 2) \right )^2 \equiv (−1)^{\frac{p+1}{2}} \pmod p $$