Final Round of National Olympiad

By manipulating the given product we get
$$\begin{aligned} \left(1 + \frac{1}{1}\right) \cdot \left(1 + \frac{1}{3}\right) \cdot \left(1 + \frac{1}{5}\right) \cdots \left(1 + \frac{1}{2n-1}\right) &= \frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdots \frac{2n}{2n-1} \\ &= \frac{2 \cdot 4 \cdot 6 \cdots 2n}{1 \cdot 3 \cdot 5 \cdots (2n-1)}. \end{aligned}$$
For it to be an integer, the number $2 \cdot 4 \cdot 6 \cdots 2n$ should be divisible by $1 \cdot 3 \cdot 5 \cdots (2n - 1)$. But since
$$2 \cdot 4 \cdot 6 \cdots 2n = (2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdots (2 \cdot n) = 2^n \cdot (1 \cdot 2 \cdot 3 \cdots n)$$
and number $1 \cdot 3 \cdot 5 \cdots (2n - 1)$ is odd, number $1 \cdot 2 \cdot 3 \cdots n$ should be divisible by $1 \cdot 3 \cdot 5 \cdots (2n - 1)$. In case of $n > 1$ this is impossible, because $1 \cdot 2 \cdot 3 \cdots n < 1 \cdot 3 \cdot 5 \cdots (2n - 1)$.