<s>[INST] Quick, self-contained way to see why $\left({{-1}\over p}\right) = 1$? Let $p$ be a prime number congruent to $1$ modulo $4$. What is a quick and self-contained way to see why$$\left({{-1}\over p}\right) = 1?$$
 [/INST] If $p$ is any odd prime, then by Wilson's Theorem: $$(p-1)!\equiv 1\cdot 2\cdots\left(\frac{p-1}{2}\right)\left(-\frac{p-1}{2}\right)\cdots (-2)(-1)\pmod{p}$$
$$\equiv (-1)^{\frac{p-1}{2}}\left(\left(\frac{p-1}{2}\right)!\right)^2\equiv -1\pmod{p}$$
If $p\equiv 1\pmod{4}$, then all the solutions of $x^2\equiv -1\pmod{p}$ are $x\equiv \pm\left(\frac{p-1}{2}\right)!\pmod{p}$.

Edit: another proof: Let $p$ be prime, $p\equiv 1\pmod{4}$ and let $g$ be a primitive root mod $p$. Then $\text{ord}_p(g)=p-1$, so $g^{\frac{p-1}{2}}\equiv -1\pmod{p}$. Therefore, all the solutions of $x^2\equiv -1\pmod{p}$ are $x\equiv \pm g^{\frac{p-1}{4}}\pmod{p}$.
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