prove limit of exponential function without concept of logarithm

Question

The question is, prove that if a real number $x>1$, then $\lim_{n\to\infty}x^n = \infty$, where $n \in \mathbb N$, without using the logarithmic concept. I came up with a proof, but I'm not so sure about it. So I want to verify if it is correct.

$n \in \mathbb N$, and define a sequence $\{x_n\}=\{x, x^2, x^3, ...\}$.

Suppose $\{x_n\}$ is bounded. By the least upper bound property of $\mathbb R$, there exists $c=\sup_n\{x_n\}$. However, since there exists an element $cx \in \{x_n\}$, and $cx>c$, it is a contradiction.

Therefore, $\{x_n\}$ is not bounded. Since it is non-empty and unbounded, for any $M\in \mathbb N$, there exists $N \in \mathbb N$ such that $x^N > M$.

Since $\{x_n\}$ is monotone increasing, for any $M\in \mathbb N$, there exists $N \in \mathbb N$ such that if $m > N$, $x^m > M$. Hence $\lim_{n\to \infty} x^n = \infty$.

Is it flawless?

Answer 1

However, since there exists an element $cx \in \{xn\}$, and $cx>c$, it is a contradiction.

This is false, and I cannot figure out how you contrived this claim. $\def\nn{\mathbb{N}}$

Instead, since $c = \sup(\{x_n:n\in\nn\}) \ge x > 1$, let $y \in \{x_n:n\in\nn\}$ such that $y>\frac{c}{x}$ since $\frac{c}{x} < c$. Then $xy > c$, which is a contradiction.

Answer 2

Here is another approach. Since $x >1$, we have $x=1+t$ for some $t>0$. Then $x^n = (1+t)^n = \sum_k \binom{n}{k} t^k \ge 1 + n t$.

Since $1+nt \to \infty$, we have the desired result.