<s>[INST] Another Congruence Proof I've been asked to attempt a proof of the following congruence.  It is found in a section of my textbook with Wilson's theorem and Fermat's Little theorem.  I've pondered the problem for a while and nothing interesting has occurred to me.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2} (\text{mod} p)$
 [/INST] Here  $p$ must be an odd prime. 
There are two cases to consider, $p\equiv 1 \pmod 4$ and $p\equiv 3 \pmod{4}$. We deal with the first case.
We know that $(p-1)!\equiv -1\pmod{p}$.  Rearrange the numbers from $1$ to $p-1$, so that we get the odd numbers from $1$ on going up, interleaved with the even numbers from $p-1$ going down.  For example, if $p=13$, we arrange the numbers from $1$ to $12$ in the order $1$, $12$, $3$, $10$, $5$, $8$, $7$, $6$, $9$, $4$, $11$, $2$. 
In general the listing is $1$, $p-2$, $3$, $p-3$, $5$, $p-5$, and so on until at the end we get to $p-2$, followed by $p-(p-2)$. Now take the product, in that order,  noting that $p-k\equiv -k \pmod{p}$. 
We get that 
$$(p-1)!\equiv (1)(-1)(3)(-3)(5)(-5)\cdots (p-2)(-(p-2))\equiv -1\pmod{p}.\tag{$1$}$$
The number of even numbered entries in the product is $(p-1)/2$. These have minus signs in front of them. Gather the minus signs together. We get
$$(-1)^{(p-1)/2} 1^23^25^2\cdots (p-2)^2 \equiv -1\pmod{p}.$$
Note that $(p-1)/2$ is even. So we get that
 $$1^23^25^2\cdots (p-2)^2 \equiv -1\pmod{p}.$$
Now we are finished, since $-1=(-)^{(p+1)/2}$.
The argument for $p\equiv 3\pmod{4}$ is essentially the same. In fact the two arguments could be gathered into one.  The main difference is that in $(1)$, the number of minus signs, which is $(p-1)/2$, now turns out to be odd. So we get
$$-1^23^25^2\cdots (p-2)^2 \equiv -1\pmod{p}.$$
or equivalently 
$$1^23^25^2\cdots (p-2)^2 \equiv 1\pmod{p}.$$
Since in this case we have $(-1){(p+1)/2}=1$, again the result follows.  
Remark: The above shows the useful result that if $p\equiv 1\pmod{4}$, then there is an $x$ such that $x^2\equiv -1\pmod{p}$. Indeed it gives an explicit expression for such an $x$. Regrettably, the expression is not computationally useful if $p$ is large.
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