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amparvardi wrote: $P$ is divisible by $pk$, then $\frac{P}{p^k} \geq n!.$ Should be $P$ is divisible by $p^k$. It suffices to show this for the greatest k such that $p^k$ divides P. Let $x_i = p^{k_i}r_i$, such that $r_i$ is not divisible by p. Note $k=k_1 + k_2 + \cdots + k_n$, and that $\frac{P}{p^k}= (r_1)(r_2)\cdots (r_n)$. Suppose that $r_a = r_b$ for some $1 \le a< b \le n$. Since $x_b > x_a$, it is also true that $p^{k_b} > p^{k_a}$. Then, $\frac{x_b}{x_a} = p^{k_b - k_a} < 2$, which is an impossibility. Therefore, all $r_i$ must be distinct, so $(r_1)(r_2)\cdots (r_n) \ge n!$

aboojiga wrote: amparvardi wrote: $P$ is divisible by $pk$, then $\frac{P}{p^k} \geq n!.$ Should be $P$ is divisible by $p^k$. Thanks, edited