To prove $n\binom{2n-1}{n-1}$ is divisible by $n(2n-1)$ 
Prove that for natural $n \ge 2$ $$n\binom{2n-1}{n-1}$$ is divisible by $$n(2n-1)$$

We have $$n\binom{2n-1}{n-1}=n \frac{(2n-1)!}{(n-1)! \:n!}=n(2n-1)\frac{(2n-2)!}{(n-1)! \:n!}$$
Now it suffices to prove $\frac{(2n-2)!}{(n-1)! \:n!}$ is an integer
Now
$$\frac{(2n-2)!}{(n-1)! \:n!}= \frac{1 \times 2 \times 3 \cdots \times (n-1) \times (n-2) \times (n-3) \cdots  \times (2n-4) \times (2n-3) \times (2n-2)}{(n-1)! \: n!}$$
hence
$$\frac{(2n-2)!}{(n-1)! \:n!}=\frac{(n-2) \times (n-3) \cdots (2n-2)}{n!}$$
any clue to prove this is always an integer?

Because $$\frac{\binom{2n-1}{n-1}}{2n-1}=\frac{\binom{2n-2}{n-1}}{n}$$ and
$gcd(2n-1,n)=1$.
