We may assume that $|P_0P_1| \le |P_0P_2| \le \dots \le |P_0P_n|$.
At first, we will prove that $|P_0P_k| > \frac{d}{3}\sqrt{k+1}$ for any positive integer $n$.
Obviously, $|P_0P_k| \ge d \ge \frac{d}{3}\sqrt{k+1}$ for $k = 1, 2, \dots, 8$, and the second equality holds only when $k = 8$. Then we only need to prove that $|P_0P_k| \ge d \ge \frac{d}{3}\sqrt{k+1}$ for $k \ge 9$.
Take each $P_i$ ($i = 0, 1, 2, \dots, k$) as the center to draw a circle with radius $\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other. Take $P_0$ as the center to draw a circle with radius $|P_0P_k| + \frac{d}{2}$. Then the previous $k+1$ smaller circles are all located in this larger one.
Then $\pi(|P_0P_k| + \frac{d}{2})^2 > (k+1)\pi(\frac{d}{2})^2$, from which we have $|P_0P_k| > \frac{d}{2}(\sqrt{k+1} - 1)$.
It is easy to check that $\frac{\sqrt{k+1}-1}{2} > \frac{\sqrt{k+1}}{3}$ for $k \ge 9$.
Then $|P_0P_k| > \frac{d}{3}\sqrt{k+1}$ for $k \ge 9$.
Over all, we have $|P_0P_k| > \frac{d}{3}\sqrt{k+1}$ for $k \ge 9$.
Therefore,
$$|P_0P_1| \cdot |P_0P_2| \cdots |P_0P_n| > \left(\frac{d}{3}\right)^n \sqrt{(n+1)!}$$

We may assume $|P_0P_1| \le |P_0P_2| \le \cdots \le |P_0P_n|$.
Take each $P_i$ ($i = 0, 1, 2, \dots, k$) as the center to draw a circle with radius $\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other.
Let $Q$ be any point on $\odot P_i$. Since
$$\begin{aligned} |P_0Q| &\le |P_0P_i| + |P_iQ| = |P_0P_i| + \frac{d}{2} \\ &\le |P_0P_k| + \frac{1}{2} |P_0P_k| = \frac{3}{2} |P_0P_k|, \end{aligned}$$
we get that the circle with center $P_0$ and radius $\frac{3}{2} |P_0P_k|$ cover the previous $k+1$ smaller circles. Then we have $\pi(\frac{3}{2} |P_0P_k|)^2 > (k+1)\pi(\frac{d}{2})^2$, and that is
$$| P_0 P_k | > \frac{d}{3} \sqrt{k+1} \quad (i = 0, 1, 2, \dots, k).$$
Therefore,
$$| P_0 P_1 | \cdot | P_0 P_2 | \cdots | P_0 P_n | > \left(\frac{d}{3}\right)^n \sqrt{(n+1)!}$$