Set $n=qp+r$ with $0\le r\le p-1$, the remainder obtained upon dividing $n$ by $p$. Then $\lfloor n/p\rfloor = q$. Now, \[ \binom{n}{p} = \frac{n!}{(n-p)!p!} = \frac{n(n-1)\cdots(n-p+1)}{p(p-1)!} \equiv \frac{n(n-1)\cdots (n-r+1)}{(p-1)!}\cdot \frac{n-r}{p} \cdot (n-r+1)\cdots(n-p+1). \]Using $(n-r)/p=q$, $(p-1)!\equiv -1\pmod{p}$ and the fact $\{n,n-1,\dots,n-r+1,n-r-1,\dots,n-p+1\}$ coincides with $\{1,2,\dots,p-1\}$ modulo $p$, we obtain the desired conclusion.

grupyorum wrote: Set $n=qp+r$ with $0\le r\le p-1$, the remainder obtained upon dividing $n$ by $p$. Then $\lfloor n/p\rfloor = q$. Now, \[ \binom{n}{p} = \frac{n!}{(n-p)!p!} = \frac{n(n-1)\cdots(n-p+1)}{p(p-1)!} \equiv \frac{n(n-1)\cdots (n-r+1)}{(p-1)!}\cdot \frac{n-r}{p} \cdot (n-r+1)\cdots(n-p+1). \]Using $(n-r)/p=q$, $(p-1)!\equiv -1\pmod{p}$ and the fact $\{n,n-1,\dots,n-r+1,n-r-1,\dots,n-p+1\}$ coincides with $\{1,2,\dots,p-1\}$ modulo $p$, we obtain the desired conclusion. I believe there is a typo in $\frac{n-r}{p} \cdot (n-r+1)\cdots(n-p+1).$ should probably have been $\frac{n-r}{p} \cdot (n-r-1)\cdots(n-p+1).$

Claim: $\binom {kp}{p} \equiv k$ mod $p$. Proof: Expand $\binom {kp}{p}$, then simplify. Now let $n = qp + r, r<p$ Claim: If $r \ne p-1,$ and $\binom {qp + r}{p} \equiv q$ mod $p$, then $\binom {qp + r + 1}{p} \equiv q$ mod $p$. Proof: $\binom {qp + r + 1}{p} = \binom {qp + r}{p} \cdot \frac{qp+r+1}{(q-1)p+r+1} \equiv q \cdot \frac{qp+r+1}{(q-1)p+r+1} \equiv q$ mod $p$. And we are done from induction. $\square$