Prove that $\forall n, \, \exists N,x :\lfloor{x^{N}}\rfloor =n \, \land \,\lfloor{x^{N+1}}\rfloor =n+1$ The question is related to the interesting problem raised by following OP.
Notably I'm trying to prove the following fact

For any $n\in \mathbb{N} \quad \exists x\in \mathbb{R} \quad x>1$ and $\exists N\in \mathbb{N}$ such that
$$\lfloor{x^{N}}\rfloor =n \quad \land \quad \lfloor{x^{N+1}}\rfloor =n+1$$

My idea for the proof is to consider

*

*$x=\sqrt[N]{n} \implies \lfloor{x^{N}}\rfloor = \lfloor{{(\sqrt[N]{n})}^{N}}\rfloor =\lfloor n\rfloor=n$
and then we need that

*

*$x^{N+1}=x\cdot n=n+1+\alpha\,$ for some $\alpha \in(0,1)$
therefore finally we need to show that for any $n\in \mathbb{N} \quad \exists \alpha\in \mathbb{R} \quad \alpha \in(0,1)$ and $\exists N\in \mathbb{N}$ such that
$$\sqrt[N]{n}=1+\frac{1+\alpha}{n}$$
but I'm totally stuck here and I can't find any method to prove that.
Thanks in advance for any idea or suggestion about that!

This is long after you received the excellent answer by Ingix(+1), but I'd like to present an arguably more elementary solution.

We'll use $x = n^{1/N}$ as in your Q. We need to argue that $\exists N \in \mathbb{N}:$ \begin{align} n+1 &\le n^{1 + 1/N} < n+2 \\ \iff \frac{\ln (n+1)}{\ln n} &\le 1 + \frac{1}{N} < \frac{\ln (n+2)}{\ln n} \\ \iff \frac{\ln(1+1/n)}{\ln n} &\le \frac{1}{N} < \frac{\ln (1 + 2/n)}{\ln n} \\ \iff \frac{\ln n }{\ln(1 + 2/n)} &< N \le \frac{\ln n}{\ln( 1 + 1/n)}
\end{align}
But this is true so long as the difference between the left and right hand side above is at least $1$. We can use the the following elementary inequality: $$\mathrm{for}\, x\in (0,1): \quad 2x/3 \le \ln(1+x) \le x \iff \frac{1}{x} \le \frac{1}{\ln(1+x)} \le \frac{3}{2x}. $$
Thus, $ \frac{\ln n}{\ln(1 + 1/n)} \ge n \ln n$ and $ \frac{\ln n}{\ln(1 + 2/n)} \le \frac{3n}{4} \ln n$. The gap between them is at least $  \frac{n}{4} \ln n $ which is seen to be $> 1$ for $n\ge 4.$
For $n = 1,2,3$ witnesses are found easily: $x = 4/3$ and $N = 2,3,4$ respectively.
