2020/028) Compare $\sqrt[n+1]{(n+1)!}$ and $\sqrt[n]{(n)!}$ where n is integer

 We have $(n+1)!= (n+1)n!$

Taking $n^{th}$ power we have $((n+1)!)^n = ((n+1)n!)^n = (n+1)^n (n!)^n\cdots(1)$
Now $(n+1)^n > n!\cdots(2)$
From (1) and (2) we get $((n+1)!)^n > (n!)^{n+1}$
Taking $(n(n+1))^{th}$ root we get $\sqrt[n+1]{(n+1)!} > \sqrt[n]{(n)!}$