${{p-1}\choose{j}}\equiv(-1)^j \pmod p$ for prime $p$ Can anyone share a link to proof of this?
$${{p-1}\choose{j}}\equiv(-1)^j(\text{mod}\ p)$$ for prime $p$.
 A: $$\binom {p-1}j=\prod_{1\le r\le j}\frac{p-r}r$$
Now, $\displaystyle p-r\equiv -r\pmod p\implies \frac{p-r}r\equiv-1\pmod p$
A: Since $p$ is prime, $\displaystyle\binom pj\equiv0\pmod p$ for $0\lt j\lt p$.  
By Pascal's identity, for $0\lt j\lt p$ we have $$\binom{p-1}j=\binom pj-\binom{p-1}{j-1}\equiv-\binom{p-1}{j-1}\pmod p.$$Since $\displaystyle\binom{p-1}0=1$, it follows by induction that $\displaystyle\binom{p-1}j\equiv(-1)^j\pmod p$ 
for $0\le j\le p-1$.
A: It is well known that $\binom pi\equiv0\pmod p$ for $0<i<p$. Now Pascal's recurrence gives $\binom{p-1}i\equiv-\binom{p-1}{i-1}\pmod p$ for those$~i$, and so $\binom{p-1}i\equiv(-1)^i\pmod p$ for $0\leq i<p$ follows by an immediate induction on$~i$, with $\binom{p-1}0=1$ as base case.
More generally this gives $\binom{p^k-1}i\equiv(-1)^i\pmod p$ for $0\leq i<p^k$ for any positive integer$~k$ (since $\binom{p^k}i\equiv0\pmod p$ for $0<i<p^k$). This result would be a bit harder to prove by just reducing modulo$~p$ all factors in the expansion of $\binom{p^k}i$ than the basic case (since now some factors in numerator and denominator reduce to$~0$ modulo$~p$).
A: Definition of $a \equiv b \pmod{c}$ requires $a,b,c$ to be integers. (See David Burton's Elementary Number Theory for a definition and a similar problem.) Here is a way to do it. $$(p-1)(p-2)\ldots(p-j) \equiv (-1)^j j! \pmod{p}.$$ Therefore, $$\binom{p-1}{j} j! \equiv (-1)^j j! \pmod{p}.$$ Now we can "cancel" $j!$ because $\gcd(j!, p)=1$ for $1\leq j \leq p-1$ to obtain the result.
