My solution: We proceed by induction. The base case is trivial. Assume the result for some natural $n$ greater than $2$. Now, consider $((n+1)!)!=(n!)!\cdot \displaystyle\prod_{i=1}^{(n-1)(n!)-1} (n!+i)\cdot \displaystyle\prod_{j=0}^{n!-1}(n\cdot n!+j)\cdot (n+1)!$ $>n[(n-1)!]^{n!}\cdot (n!)^{(n-1)(n!)-1}\cdot (n\cdot n!)^{n!}\cdot (n+1)!$ $=n\cdot [(n-1)!]^{n!}\cdot (n!)^{(n-1)(n!)-1}\cdot n^{n!}\cdot (n!)^{n!}\cdot (n+1)\cdot n!$ $>[(n-1)!\cdot n]^{n!}\cdot (n!)^{(n-1)(n!)-1+n!+1}\cdot (n+1)$ $=(n+1)\cdot (n!)^{n!}\cdot (n!)^{n(n!)}$ $=(n+1)\cdot (n!)^{(n+1)!}$ as required.