TITLE: $n+1$ is a divisor of $\binom{2n} {n}$
QUESTION [7 upvotes]: I want to show that $n+1$ is a divisor of $\displaystyle\binom{2n} {n}$, for all $n\in\mathbb{N}.$
I have tried to show it by induction and Pascal's rule but it did not worked out.
I would appreciate some help.

REPLY [5 votes]: Hint $\ $ Put $\rm\:k = 2n\:$ in the following
Theorem $\displaystyle \rm\ \ (n\!+\!1,k\!+\!1) = 1\:\Rightarrow\: n\!+\!1\:\bigg|\:{k\choose n}$
Proof $\displaystyle\rm\ \ \ (n\!+\!1){k\choose n\!+\!1}\, =\, \frac{(n\!+\!1)\,k!}{(n\!+\!1)! \,(k\!-\!n\!-\!1)!}\, =\, \frac{(k\!-\!n)\,k!}{n!\, (k\!-\!n)!} \,=\, (k\!-\!n){k\choose n}$
so the result follows from Euclid's Lemma, since $\rm\:(n\!+\!1,k\!-\!n) = (n\!+\!1,k\!+\!1) = 1.\ \ $ QED