If $\lfloor x^i\rfloor =i,i=1,2,3,\cdots,n$ find the maximum of $n$

Question

Find the maximum $n$ for which there exist a real number $x$ such that $$\lfloor x^i\rfloor =i,\quad i=1,2,3,\ldots,n.$$

$\lfloor x\rfloor =1$,then $1<x<2$,

$\lfloor x^2\rfloor =2$ then $\sqrt{2}<x<\sqrt{3}$. so How to find the maximum of $n?$

Answer 1

Hint: $(\root3\of3)^5>6.{}$