Let $p_i $ be a prime divisor of $k $. Since $p_i\le \frac{k}{2}<k$ we get $p_i\mid (k-1)! $. Let $\nu_{p_i}(k)=u_i $. Thus since $p_i\ge 2$ we have $ \sum_{l=1}^\infty\left\lfloor \frac {k-1}{p_i^l}\right\rfloor\ge u_i $ and so $\gcd ((k-1)!,k)=k $. Similarly with $k+1$ and $k+2$ with minor work in $\nu_2, \nu_3$.