What is the smallest real number $x$ such that $\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor$ for all positive integers $n$?
In particular, we have $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<\dots$. For the first inequality to hold it must be that $x\geq\sqrt{2}$. But if $x=\sqrt{2}$ then $\lfloor x^2\rfloor=\lfloor x^3\rfloor=2$, so we need $x\geq \sqrt[3]{3}\approx 1.44$. From the first few powers it looks like this is the answer, but how can we prove it?
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Let $x=\sqrt[3]{3}$. First, you've already directly verified that $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor$, as they are equal $1$, $2$, and $3$, respectively. Now, for all $n\ge3$: $$x^{n+1}-x^n=x^n(x-1)=\left(\sqrt[3]{3}\right)^n\left(\sqrt[3]{3}-1\right)>3\cdot0.4>1,$$ so the numbers grow by more than $1$ each time, guaranteeing that their integer parts strictly increase.
You have correctly shown that $x \ge \sqrt[3]3.\ $ Now if $x^n \ge 3$ and $x \gt \sqrt [3]3, \lfloor x^{n+1}\rfloor= \lfloor x^n\cdot x\rfloor\ge\lfloor x^n\rfloor+\lfloor x^n(x-1)\rfloor \ge \lfloor x^n \rfloor +1$ so any higher $x$ will work as well, so $\sqrt[3]3$ is the smallest $x$