Proving $\prod_{j=1}^n \left(4-\frac2{j}\right)$ is an integer

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*}

Using Pascal identity

$$\displaystyle\binom{2n}{n}=\left(4-\frac{2}{n}\right)\cdot\binom{2n-2}{n-1}$$

one gets for your product

$$\displaystyle\frac{\binom{2}{1}}{\binom{0}{0}}\cdot\frac{\binom{4}{2}}{\binom{2}{1}}\cdots\frac{\binom{2n-2}{n-1}}{\binom{2n-4}{n-2}}\cdot\frac{\binom{2n}{n}}{\binom{2n-2}{n-1}}=\binom{2n}{n}$$

and this is an integer as required.