<s>[INST] Floor functions of powers increase 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?
 [/INST] 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.
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