${{p-1}\choose{j}}\equiv(-1)^j \pmod p$ for prime $p$ Can anyone share a link to proof of this?
$${{p-1}\choose{j}}\equiv(-1)^j(\text{mod}\ p)$$ for prime $p$.

$$\binom {p-1}j=\prod_{1\le r\le j}\frac{p-r}r$$
Now, $\displaystyle p-r\equiv -r\pmod p\implies \frac{p-r}r\equiv-1\pmod p$
