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.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2} (\text{mod} p)$
 A: HINT: For $p=11$:
$$\begin{align*}
10!&=(1\cdot10)(2\cdot9)(3\cdot8)(4\cdot7)(5\cdot6)\\
&=\Big(1\cdot(-1)\Big)\Big(-9\cdot9\Big)\Big(3\cdot(-3)\Big)\Big(-7\cdot7\Big)\Big(5\cdot(-5)\Big)\\
&=(-1^2)(-9^2)(-3^2)(-7^2)(-5^2)\\
&=(-1)^5\cdot1^2\cdot3^2\cdot5^2\cdot7^2\cdot9^2
\end{align*}$$
A: 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.
A: To Prove : $1^2.2^2.3^2....(p-1)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
We know that
$$k \equiv -(p-k) \pmod p$$
Applying this repeatedly we will get
$$2.4.6.......(p-1) \equiv (-1)^{\frac{p-1}{2}} 1.3.5.......(p-2) \pmod p$$
$$\implies 1.3.5.......(p-2) \equiv (-1)^{\frac{p-1}{2}} 2.4.6.......(p-1) \pmod p$$
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} 1.2.3.4.5.6.......(p-1) \pmod p$$
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} (p-1)! \pmod p$$
By Wilson's Theorem
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} (-1) \pmod p$$
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
A: An idea that, perhaps, will appeal to you besides the ones already given above:$$1^23^2\cdot\ldots\cdot (p-1)^2=\frac{\left(1\cdot 2\cdot\ldots\cdot (p-1)\right)^2}{\left(2\cdot 4\cdot\ldots\cdot (p-1)\right)^2}\,\,(**)$$Now we use Wilson's theorem, Fermat's Little Theorem and arithmetic modulo $p$: $$(**)\,\,=\frac{(-1)^2}{\left(2^{\frac{p-1}{2}}\right)^2\left(1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\right)^2}=\frac{1}{1\cdot\left(1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\right)^2} \,\,\,(***)$$
Let us now check more closely Wilson's theorem (again, arithmetic modulo $p$):$$-1=1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\ldots\cdot (p-1)=$$$$=1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\left(-\frac{p-1}{2}\right)\cdot\ldots\cdots (-2)(-1)=$$$$=\left(-1\right)^{\frac{p-1}{2}}\left(1\cdot 2\cdot\ldots\cdot\frac{p-1}{2}\right)^2$$So: $$(1)\,\,p=3\pmod 4\Longrightarrow \frac{p-1}{2}\text{ is odd}\Longrightarrow (-1)^\frac{p-1}{2}=-1\Longrightarrow $$$$\Longrightarrow\,\,(***)=1$$$$(2)\,\,p=1\pmod 4\Longrightarrow \frac{p-1}{2}\text {is even}\,\Longrightarrow (-1)^{\frac{p-1}{2}}=1\Longrightarrow$$$$\Longrightarrow (***) =-1$$
