Computing the value of $(\frac{p-1}{2})!$ modulo $p$. I want to prove that for $p \geq 3$, and for $a=(\frac{p-1}{2})!$, if $p \equiv1\pmod 4$, then $a^2\equiv -1 \pmod p$, and if $p \equiv 3\pmod4$, then $a \equiv +/-1 \pmod p$.
For the first part, I used Wilson's theorem which says that for prime $p$, $(p-1)!=-1 \pmod p$. 
so $a^2=((\frac{p-1}{2})!)^2=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))^2$ and since $p-k \equiv -k \pmod p$ we get that $a^2=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))=(1\cdot2\cdot3\cdot\cdot\cdot(\frac{p-1}{2}))((p-1)(p-2)(p-3)\cdot\cdot\cdot(-1)^{(p-1)/2}).$ 
so since $p \equiv 1\pmod 4$ and using Wilson I get the what I need. Is this correct? How should I prove the second part? 
(what is the latex for +/- symbol?)
Thank you.

You got the hard part. To summarize, we have $a^2\equiv (-1)^{(p+1)/2}\pmod{p}$.
If $p$ is of the form $4k+1$, then $(p+1)/2$ is odd, and therefore $a^2\equiv -1\pmod{p}$.
If $p$ is of the form $4k+3$, then $(p+1)/2$ is even, and therefore $a^2\equiv 1\pmod{p}$.  It follows that $a\equiv 1\pmod{p}$ or $a\equiv -1\pmod{p}$. 
This is because from $p$ divides $a^2-1$, we see that $p$ divides $(a-1)(a+1)$. Therefore either $p$ divides $a-1$, in which case $a\equiv 1\pmod{p}$ or $p$ divides $a+1$, in which case $a\equiv -1\pmod{p}$.
Remark: Note that for example when $p=3$, then $a=1$, so $a\equiv 1\pmod{p}$.  It is also the case that when $p=23$, we have $a\equiv 1\pmod{p}$. But, for example, when $p=7$, and when $p=11$, we have $a=\equiv -1\pmod{p}$. So for $p$ of the form $4k+3$, both $a\equiv 1\pmod{p}$ and $a\equiv -1\pmod{p}$ can occur.  Whether $a$ in this case is congruent to $1$ or $-1$ turns out to be connected with the solvability of the congruence $x^2\equiv 2\pmod{p}$.
