Show that $2n\choose n$ is divisible by $2n-1$ I have found many questions asking to prove $2n\choose n$ is divisible $2$, but I also observed by trying the first few "$n$" that $2n\choose n$ is divisible  by $2n-1$. 
It sure seems true when $2n-1$ is prime, but is it true in general for all $n$?

Note that $\dbinom{2n}{n}$ is an integer. Note also that $1$ divides $\binom  21 $ and assume $n>1$ hereafter.
$\begin{align}
\dbinom{2n}{n} &= \dfrac{2n!}{n!\cdot n!} \\ &= \dfrac{2n\cdot (2n-1)\cdots (n+1)}{n\cdot (n-1)\cdots 1}\\
&= \dfrac{(2n)(2n-1)}{n\cdot n}\dbinom{2(n-1)}{n-1}\\
&= 2\dfrac{2n-1}{n}\dbinom{2(n-1)}{n-1}\\
\end{align}$
Now $\gcd(2n-1,n)=1$ so $n$ must divide  $2\dbinom{2(n-1)}{n-1}$, that is, $2\dbinom{2(n-1)}{n-1}= kn$ for some integer $k$.
Then $\dbinom{2n}{n} = k(2n-1)$ as required.
