Clearly $x>1$. And for $1<x< \sqrt[3]{3}$ we know two of $\lfloor x \rfloor$ , $\lfloor x^2 \rfloor$ and $\lfloor x^3 \rfloor$ are equal. Thus $x= \sqrt[3]{3}$. Easily we check: $\lfloor \sqrt[3]{3} \rfloor < \lfloor \sqrt[3]{3^2} \rfloor < \lfloor \sqrt[3]{3^3} \rfloor <\lfloor \sqrt[3]{3^4} \rfloor <\lfloor \sqrt[3]{3^5} \rfloor$. And for $n \ge 5$ $ \sqrt[3]{3^n} +1 < \frac{7 \sqrt[3]{3^n} }{5} <\sqrt[3]{3^{n+1} }$. Thus $\lfloor \sqrt[3]{3} \rfloor < \lfloor \sqrt[3]{3^2} \rfloor < \lfloor \sqrt[3]{3^3} \rfloor \cdot \cdot \cdot <\lfloor \sqrt[3]{3^n} \rfloor <\lfloor \sqrt[3]{3^{n+1} } \rfloor$.

GovernmentCheeseAhoy wrote: Find the minimum real $x$ that satisfies $$\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<...<\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor<...$$

always_correct wrote: GovernmentCheeseAhoy wrote: Find the minimum real $x$ that satisfies $$\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<...<\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor<...$$

You are not always correct $\lfloor 2 \rfloor = \lfloor 2 \sqrt{2} \rfloor =2$

My bad, I guess only proved $\sqrt{2}$ was a lower bound.

$x= \sqrt[3]{3}$. If $x<\sqrt[3]{3}$ then $x^3<3$. Lets prove $3^x\ge x^3$ by induction. First, we do $x=1,2,3$ and notice it stands. Suppose it does for $x>3$. $3^{x+1}>3x^3>(x+1)^3=x^3+3x^2+3x+1$. So we must prove $2x^3-3x^2-3x-1>0$. Let $f(x)=2x^3-3x^2-3x-1$, then $f'(x)=3(2x^2-2x-1)$, $f'(x)=0$ for $x=\frac{1\pm \sqrt 3}{2}$. Our functions slope becomes positive for $x>\frac{1+ \sqrt 3}{2}$. $f(3)>0$, so $f(x)>0$ for $x>3$, and we already saw that $x=1, 2$ and $3$ do work. So we are finish.

Why don't you fix the question in latex ?

GovernmentCheeseAhoy wrote: Find the minimum real x that satisfies [x]<[x^2]<[x^3)<...<[x^n]<[x^n+1]<... We see that $x > 1$ and $x^n \geq n$ is needed (as per the conditions of the problem) and therefore $x \geq n^{1/n}$ where $n$ is positive integer exceeding $1$ and maximum of $n^{1/n}$ over $\mathbb{N}_{\geq 2}$ is $3^{1/3}$ (mainly because over $\mathbb{R}_{>1}$ the maxima of $x^{1/x}$ occurs at $x = e$ and looking at values and the graphs it can be seen that over $\mathbb{N}_{\geq 2}$ the maxima must occur at $x = 2$ or $x = 3$, the maxima occurring at latter) and therefore $x \geq \sqrt[3]{3}$. Easy to show $x = \sqrt[3]{3}$ indeed works

We claim that the answer is $\sqrt[3]{3}.$ Indeed, no smaller number works: It is clearly impossible to have $x\leq 1$, so $x>1.$ Since $\lfloor x \rfloor , \lfloor x^2 \rfloor$ and $\lfloor x^3 \rfloor$ must be distinct, we have $$\lfloor x^3 \rfloor \geq 3 \Rightarrow x^3 \geq 3.$$Note that for $x=\sqrt[3]{3}$, the inequalities $\lfloor x^{k+1} \rfloor > \lfloor x^{k} \rfloor$ are true for $k\leq 2$ and for $k>2$, we have $\lfloor x^{k} \rfloor \geq 3.$ We then deduce from $x > \frac{4}{3}$ that $$x^{k+1} \geq x \lfloor x^k \rfloor > \frac{4}{3} \lfloor x^k \rfloor \geq \lfloor x^k \rfloor+1,$$and so $\sqrt[3]{3}$ actually works.