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.

We could cheat and prove a more general result. Suppose that for all $k$ with $-\le k\le n$, the binomial coefficient $\binom{n}{k}$, or more properly $\frac{n!}{k!(n-k)!}$ is an integer. Then from the easily verified "Pascal" identity
$$\binom{n+1}{j}=\binom{n}{j}+\binom{n}{j-1},$$
the desired result holds for $\binom{n+1}{j}$ for all $j$ with $0\le j\le n+1$. 
Of course we should not use binomial coefficient notation, and should instead manipulate factorials. So the Pascal identity should be proved the clumsy way, not by using its definiton in terms of "choosing." 
