Question 41
Prove that the number $A = \frac{(4n)!}{n!\,(2n)!}$, where $n$ is a positive integer, is an integer and has a factor of the form $2^{n+1}$. (Note: The number $n!$ for $n \in \mathbb{N}$, is defined by: $n! = 1 \cdot 2 \cdot \dots \cdot n$, and $0! = 1$.)