Proving that (4-2/1)(4-2/2)...(4-2/n) in an integer. We have to prove that $(4-2/1)(4-2/2)...(4-2/n)$ is an integer for $n\in\mathbb{N}$. Can we do this by induction?

We prove for $n = 1$, which is trivial as $(4-2/1) = 2$ which is clearly an integer.

next we assume that the statement is true for some integer $n$
i.e. that $(4-2/1)(4-2/2)...(4-2/n) = p$ where p is an integer, and we prove for $n+1$.

So we need to show that $(4-2/1)(4-2/2)...(4-2/n)(4-2/(n+1))$ is also an integer. Which is the same as saying $p\times(4-\frac{2}{n+1})$ is an integer. 

We can rewrite $p$ as $\dfrac{2^n(1\times3\times5\times...\times(2n-1))}{n!}$ so essentially all we need to prove is that $n+1$ divides the numerator of $p$. So assume that $n$ is even, which implies that $n+1$ is odd and certainly less than $2n-1$ so therefore it will divide one of the odd numbers in the numerator of $p$. If $n$ is odd, then $n+1$ is even and will not have more factors of $2$ than $2^n$. So if the prime factorization of $n+1$ contains $k$ factors of 2, $k&lt;n$ or $k=n$,  implying that $2^k$ divides $2^n$. Also the prime factorisation of $n+1$ will not have have an odd factor that is greater than $2n-1$ as it already has a factor of 2. Therefore $n+1$ divides the numerator of p. 

Can we do this? It seems like it is not entirely correct! I may have made a big mistake in my reasoning. 

Any help is appreciated.
We can write $p$ on this form

$$p=\frac{2^n(1\times 3\times 5\times \cdots\times(2n-1))}{n!}=\frac{(2n)!}{n!n!}={2n\choose n}\leftarrow \text{integer}$$
The product is
$$ 2 {{2n-1}\choose {n-1}} $$
which is an integer for integer $n$.

This is also equal to 
$$ 2n \choose n$$
as pointed out by another member earlier.