Let $i$ be the least non-negative integer satisfying $i\equiv n\pmod{p}$. Then, we have$$\left\lfloor{\frac{n}{p}}\right\rfloor = \frac{n-i}{p}\qquad (0\le i < p)$$

We also note because $n\equiv i\pmod{p}$, it must hold that $$\begin{array}{rcllcll} n-1 &\equiv& i-1& & &\pmod{p}\\ n-2 &\equiv& i-2& & & \pmod{p}\\ &\cdots\\ n-i+1 &\equiv& 1& & &\pmod{p}\\ n-i-1 &\equiv& -1&\equiv &p-1& \pmod{p}\\ n-i-2 &\equiv& -2&\equiv& p-2& \pmod{p}\\ &\cdots\\ n-p+1 &\equiv& i+1& & & \pmod{p} \end{array} $$

It follows that $$\begin{align*} &\quad n(n-1)\dots(n-i+1)(n-i-1)\dots(n-p+1)   \\ \equiv &\quad i(i-1)\dots1\cdot(p-1)\dots(i+1)  \\ \equiv &\quad (p-1)! \pmod{p} \end{align*}$$

Therefore, $$\begin{align*} C_n^p = &\quad \frac{n(n-1)\dots(n-i+1)(n-i)(n-i-1)\dots(n-p+1)}{p!}\\ \equiv &\quad n(n-1)\dots(n-i+1)(n-i-1)\dots(n-p+1)\frac{n-i}{p!} \\ \equiv &\quad (p-1)! \frac{n-i}{p!} \\ \equiv &\quad \frac{n-i}{p} \\ \equiv &\quad\left\lfloor{\frac{n}{p}}\right\rfloor \pmod{p} \end{align*}$$