4.38. Find the highest power of two that divides the number $(n+1)(n+2) \cdot \ldots \cdot 2 n$.

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4.38. Let $(2 n-1)!!=1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot(2 n-1)$ and $(2 n)!!=2 \cdot 4 \cdot 6 \cdot \ldots \cdot 2 n$. It is clear that $(2 n)!=(2 n-1)!!(2 n)!!$ and $(2 n)!!=2^{n} n!$. Therefore, $(n+1)(n+2) \times \ldots$ $\cdots \times 2 n=\frac{(2 n)!}{n!}=\frac{(2 n-1)!!(2 n)!!}{n!}=2^{n}(2 n-1)!!$, where the number $(2 n-1)!!$ is odd.