@G.K.T. Good doubt.Take some large $n$. For $n^{\sqrt n}$ we multiply $n$ $\sqrt n$ times, i.e., $\sqrt n$, $n$ times. For $n!$, we have $n- \sqrt n$ values which are larger than $\sqrt n$ and one value equal to $\sqrt n$ and $\sqrt n - 1$ values smaller than $\sqrt n$ in the multiplication part. If we consider the first 2 parts we get ${\sqrt n} ^ {n - \sqrt n + 1}$. Now, we have $\sqrt n$ terms smaller than $\sqrt n$ and all $n - \sqrt n$ terms were larger. Since $\sqrt n <<< n - \sqrt n$, this means asymptotically, $n! > n^{\sqrt n}.$