If $\,x>1$, then $\lim\limits_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x$.

Since $y-1< \lfloor y\rfloor\le y$, for every $y\in\mathbb R$, then $$ \frac{x^{n+1}-1}{x^{n}}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}< \frac{x^{n+1}}{x^n-1}, $$ and hence $$ x-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}<x+\frac{x}{x^n-1}, $$ or $$ -\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}-x<\frac{x}{x^n-1}. $$ Since both $ -\frac{1}{x^n},\,\frac{x}{x^n-1}\to 0, \quad\text{as}\quad n\to\infty, $ then $\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}\to x$.

Hint we have $$y-1\le\left\lfloor y \right\rfloor\le y$$ then use the squeeze theorem.