Proof. Let $i \ge 0$ . Then the set of numbers between 1 and $\left\lfloor \frac{n}{p^i}\right\rfloor$ is$$\left\{kp, k \ge 1, k \le \left\lfloor\frac{n}{p^{i+1}}\right\rfloor\right\}$$ This is true for every integer $i$ with $p^{i+1} \le n$ . So we have \begin{equation} \label{F1} \frac{\left\lfloor \frac{n}{p^i}\right\rfloor!}{\left\lfloor \frac{n}{p^{i+1}}\right\rfloor p^{\left\lfloor\frac{n}{p^{i+1}}\right\rfloor}} =\pfac{\left\lfloor\frac{n}{p^i}\right\rfloor}. \end{equation}Multiplying all terms with $0 \le i \le d$ , where $d$ is the largest power of $p$ not greater than $n$ , the statement follows from the generalization of Anton's congruence.