Is there a proof that $\ 0<\alpha<1\implies\alpha^n\to\ 0\ $ without alluding to numbers greater than $\ 1,\ $ other than $n$ itself? The comment in the answer to this question: https://math.stackexchange.com/q/211846 says that we can use Bernoulli's inequality to show that, not only that $\ \alpha>1\implies\lim_{n\to\infty}\alpha^n = +\infty,\ $ but also that  $\ 0<\alpha<1\implies\alpha^n\to\ 0. $
I think one way to prove the latter is the following. Let $\ 0<\alpha<1\ $ and let $\ \varepsilon>0.\ $ Then $\ \frac{1}{\alpha}>1;\ \frac{1}{\varepsilon}>0,\ $ and by Bernoulli's inequality along with the Archimedean property of real numbers, $\ \exists N\ $ such that $\ \left(\frac{1}{\alpha}\right)^n \geq \frac{1}{\varepsilon}\quad \forall n\geq N,\ \implies\ \alpha^n \leq \varepsilon\quad \forall n\geq N\ $ and we are done. My question is,

Is there an alternative proof to this (in particular,
one using a variant of Bernoulli's inequality) without alluding to
numbers greater than $\ 1,\ $ other than the fact that $\ n\geq 1\ $ as we take the limit?

The version of Bernoulli's inequality that says that $\ (1+x)^r\geq 1+rx\ $ for every real numbers $\ r\geq 1\ $ and $\ x\geq -1\ $ doesn't seem to help here.
 A: This answer doesn't use Bernoulli's inequality, but it also makes no reference to numbers greater than $1$.
Since $0<\alpha<1$, the sequence $\alpha^n$ is strictly decreasing. Moreover, it is bounded below as $\alpha^n$ is positive for all $n\in\Bbb{N}$. Let $S=\{\alpha^n:n\in\Bbb{N}\}$. By the monotone convergence theorem,
$$
\lim_{n \to\infty}\alpha^n=\inf S \, .
$$
All that is left to show is that $0=\inf S$. Clearly, $0$ is a lower bound of $S$. Moreover, if $x>0$, then $n>\log_{\alpha}(x)\implies \alpha^n<x$ as the function $\alpha^y$ is strictly decreasing. Hence, $0$ is the greatest lower bound of $S$, and the result follows.
A: I've found another convincing argument that doesn't use logs, so will continue from Joe’s answer after this line:

All that is left to show is that $0=\inf\{S\}$.

Let $\ (a_n) = \alpha^n.\ $ Then, $\ \frac{a_{n+1}}{a_n} = \alpha\ \implies\ a_{n+1} = \alpha\ a_n.\ $ Now suppose $\ \inf\{S\}=x>0.\ $ Then by definition of convergence towards $\ x,\ \exists i\in\mathbb{N}\ $ such that $\ x\leq a_i< \alpha x,\ $ which implies that $\ a_{i+1}=\alpha\ a_i<x,\ $ contradicting $\ x\ $ being the infimum of $\ S.$
Edit: apparently, a more standard proof that you might find in an elementary analysis book is the following:
When $0\lt a\lt 1$ the sequence $a^n$ is decreasing, since  $\frac{a^{n+1}}{a^n}=a\lt 1$ which is the same as $a^{n+1}\lt a^n$. As the sequence is bounded below, it has a limit, say $l$. Hence, $\lim a^{n+1}= a\lim a^n$, or $l=al$, or $(1-a)l=0$. As $a$ is not equal to $1$, we conclude that $l=0$. From this it also follows that the lemma holds for $-1\lt a \le 0$ as well (check this).
