<s>[INST] Proving $\prod_{j=1}^n \left(4-\frac2{j}\right)$ is an integer 
  
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*How do I show that the product $$\biggl(4 - \frac21\biggr) \cdot \biggl(4 - \frac22\biggr) \cdot \biggl(4 - \frac23\biggr) \cdots \biggl(4 - \frac2{n}\biggr)$$ is an integer for any $n \in \mathbb{N}$.
  

Source: www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf
 [/INST] Show that it's equal to $ { 2n \choose n } . $
Edit: Okay, here's how it goes: 
\begin{align*}
\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) 
&=\small \frac{(4 \cdot 1 - 2)}{1}
\frac{(4 \cdot 2 - 2)}{2}
\frac{(4 \cdot 3 - 2)}{3} \cdots \frac{(4n - 2)}{n}
 \\ &= \small\frac{2^n (2 \cdot 1 - 1)(2 \cdot 2 - 1)( 2 \cdot 3 - 1) \cdots (2n-1)}{n!}
\\ &= \frac{2^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}
\\ &= \frac{2^n \cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdots (2n-1)(2n)}{n! \cdot 2 \cdot 4 \cdots (2n)}
\\ &= \frac{2^n (2n)!}{ 2^n (n!)^2} = { 2n \choose n}.
\end{align*}
</s>