Where do the two $a!$s come from?

Question

I have \begin{align*} (2a)! &\equiv a! (-a) \dotsm (-3)(-2)(-1) \pmod p\\ & \equiv (-1)^a a!a!\pmod p\\ &\equiv (-1)^a a!^2\pmod p. \end{align*} The $(-1)$ is just to get the parity right and can be neglected in solving my problem, which is: from where do the two $a!$s come from? I know where one comes from but not the other. If it is from $(2a)!$ how so? I tried to send a scanned page from "An adventurers guide to Number Theory" page 172 but could not do it using your format.

Answer 1

If there is an extra condition $p=2a+1$ like what @Max said above, then

$$\begin{align*} 2a =& p-1, &2a \equiv& -1\pmod{p}\\ 2a-1=&p-2, &2a-1 \equiv& -2 \pmod{p}\\ \vdots&&\vdots&\\ 2a-(a-1)=&p-a, &a+1 \equiv& -a \pmod{p} \end{align*}$$

So $$\begin{align*} (2a)! \equiv& a!\ (a+1)\cdots(2a-1)(2a) &\pmod{p}\\ \equiv& a!\ (-a)\cdots(-2)(-1)&\pmod{p}\\ \equiv& a!\ (-1)a\cdots(-1)2(-1)1&\pmod{p}\\ \equiv& a!\ \left[(-1)^aa!\right]&\pmod{p} \end{align*}$$