1708\hspace{-16}$We Know that $\mathbf{n!=1.2.3........n=n.(n-1).(n-2)......3.2.1}$\\\\ So $\mathbf{(n!)^2=\prod_{k=1}^{n}k\times \prod_{k=1}^{n}k=\prod_{k=1}^{n}k\times \prod_{k=1}^{n}\left(n-k+1\right)=\prod_{k=1}^{n}\left(nk-k^2+k\right)}$\\\\\\ and here $\mathbf{1\leq k\leq n}$\\\\ So $\mathbf{(k-1)\geq 0}$ and $\mathbf{\left(n-k\right)\geq 0}$\\\\ So $\mathbf{(k-1).(n-k)\geq 0}$\\\\ So $\mathbf{nk-k^2+k\geq n}$\\\\ So $\mathbf{(n!)^2=\prod_{k=1}^{n}\left(nk-k^2+k\right)\geq \prod_{k=1}^{n}\left(n\right)}$\\\\ So $\mathbf{(n!)^2\geq n^n}$
