parmenides51 wrote: Prove that for any positive integer $n$, $2 \cdot \sqrt3 \cdot \sqrt[3]{4} ,,, \sqrt[n-1]{p} > n$ I assume that it is $2 \cdot \sqrt3 \cdot \sqrt[3]{4} \ \ldots \sqrt[n-1]{n} > n$ and also $n\ge 3$. Then, by induction on $n$, when $n=3$, it is obvious. Assume that for $n=k$, the inequality holds, then when $n=k+1$, \[2 \cdot \sqrt3 \cdot \sqrt[3]{4} \ \ldots \sqrt[n-1]{n} \cdot \sqrt[n]{n+1} > n \sqrt[n]{n+1} >n+1 \iff n+1 > \left(1+\frac{1}{n}\right)^n\]which is true since we know the RHS converges to $e$ when $n$ approaches infinity but LHS will grow larger and so bigger than RHS. $\blacksquare$