Induction proof: $\dbinom{2n}{n}=\dfrac{(2n)!}{n!n!}$ is an integer. Prove using induction: $\dbinom{2n}{n}=\dfrac{(2n)!}{n!n!}$ is an integer.
I tried but I can't do it.

Here I've assumed that $\binom{2n}{n+1}$ is an integer for the proof...
We induct on two things: $1)$ $n+1 \mid \binom{2n}{n}$ and $2)$ $\binom{2n}{n}$ is an integer.
These are both easily checked for $n=1,$ so now we assume it holds for $n-1$. 
Then for $\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2n-2)!(2n-1)(2n)}{(n-1)!(n-1)! n^2} = \binom{2n-2}{n-1}\frac{2(2n-1)}{n},$ we have by the inductive hypothesis that $n$ divides $\binom{2n-2}{n-1},$ so we are done. Now it suffices to show that $n+1 \mid \binom{2n}{n},$ but this follows from $\frac{n}{n+1}\binom{2n}{n} = \binom{2n}{n+1},$ so we are done.
