Peter wrote: Show that for every natural number $ n$ the product \[ \left( 4 - \frac {2}{1}\right) \left( 4 - \frac {2}{2}\right) \left( 4 - \frac {2}{3}\right) \cdots \left( 4 - \frac {2}{n}\right) \] is an integer. We have that \[ \prod_{i = 1}^{n}{(4 - \frac {2}{i})} = \frac {\prod_{i = 1}^{n}{2(2i - 1)}}{n!} = \frac {2^{n}(2n - 1)!!}{n!} = \frac {(2n)!}{n!n!} = C_{2n}^{n} \] Where $ C_{2n}^{n}$ is a binomial coefficient and, as we know, it is a positive integer.

You can also say that $\frac{\left( 4-\frac{2}{1}\right)\left( 4-\frac{2}{2}\right)\left( 4-\frac{2}{3}\right)\cdots\left( 4-\frac{2}{n}\right)}{n+1}$ is an integer because it is the $n^{th}$ catalan number. Some more proofs of it may be seen in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=349870

A55