Slovenian Mathematical Olympiad, 2008

Let us compare the expressions $\frac{\log_{10} 2 \log_{10} 3 \cdots \log_{10} (n-1)}{10^{n-2}}$ and $\frac{\log_{10} 2 \log_{10} 3 \cdots \log_{10} n}{10^{n-1}}$. The inequality
$$\frac{\log_{10} 2 \log_{10} 3 \cdots \log_{10} (n-1)}{10^{n-2}} \ge \frac{\log_{10} 2 \log_{10} 3 \cdots \log_{10} n}{10^{n-1}}$$
holds if and only if $1 \ge \frac{1}{10} \log_{10} n = \log_{10} \sqrt[10]{n}$, which is equivalent to $10 \ge \sqrt[10]{n}$ and $10^{10} \ge n$. This implies
$$\begin{aligned} \frac{\log_{10} 2}{10} &> \frac{\log_{10} 2 \cdot \log_{10} 3}{10^2} > \dots > \frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} (10^{10} - 1)}{10^{10^{10}-2}} \\ &= \frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} (10^{10})}{10^{10^{10}-1}} \end{aligned}$$
and
$$\begin{aligned} \frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} (10^{10})}{10^{10^{10}-1}} < & \frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} (10^{10} + 1)}{10^{10^{10}}} \\ < & \frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} (10^{10} + 2)}{10^{10^{10}+1}} < \dots \end{aligned}$$
So, the expression $\frac{\log_{10} 2 \log_{10} 3 \cdots \log_{10} n}{10^{n-1}}$ has the smallest possible value when $n = 10^{10} - 1$ or $n = 10^{10}$. For these two values of $n$ the value of the expression is
$$\frac{\log_{10} 2 \log_{10} 3 \cdots \log_{10} 10^{10}}{10^{10^{10}-1}}$$