If $n\ge2$, Prove $\binom{2n}{3}$ is even. Any help would be appreciated. I can see it's true from pascal's triangle, and I've tried messing around with pascal's identity and the binomial theorem to prove it, but I'm just not making any headway.

(for $n\gt2$) we have the numerator (after the initial cancellation of $(2n-3)!)$ is at least divisible by $8$, however the denominator can at most be divisible by $2$.
