jmc

We can write the given equation as $${\log }_{10}(k-2)!+{\log }_{10}(k-1)!+{\log }_{10}100={\log }_{10}(k!{)}^2.$$Then $${\log }_{10}[100(k-2)!(k-1)!]={\log }_{10}(k!{)}^2,$$so $100(k-2)!(k-1)!=(k!{)}^2.$ Then $$100=\frac{k!\cdot k!}{(k-2)!(k-1)!}=k(k-1)\cdot k=k^3-k^2.$$So, $k^3-k^2-100=0,$ which factors as $(k-5)(k^4+4k+20)=0.$ The quadratic factor has no integer roots, so $k=\boxed{5}.$