Prove existence of $n$ such that $x>(1-\epsilon)^n$ for some $x,\epsilon$ I am stuck in an analysis problem where I have to prove this inequality-
$$x&gt;(1-\epsilon)^n$$ for some $n\in\mathbb{N}$ where $x\le1$ and $\epsilon\in(0,1)$ are specified.
Background: I am solving an analysis problem where I'm supposed to show, $\sup S_x=1$ where $S_x=\{x,x^{(1/2)},...,x^{(1/n)},...\}$ when $0&lt;x\leq 1$ and $x\in\mathbb{R}$.
I have no idea how to proceed. Please help me.
We have $0\lt 1-\epsilon \lt 1$
Then $\frac 1{1-\epsilon} \gt 1$
Let $\frac 1{1-\epsilon}=1+a $  where $a\gt 0$
Then $(1+a)^n \gt na $ which implies
$\frac 1{(1+a)^n}=(1-\epsilon)^n \lt \frac 1{na} \to 0$ as $n\to \infty$
So whatever $x$ is given in $(0,1)$ , there exist some $n\in \mathbb{N}$ such that
$(1-\epsilon)^n \lt x$
Edit : If you don't want to use limits, note that by Archimdean Property of Real numbers, there is $n_o\in \mathbb{N} $ such that
$n_o(xa)\gt 1$ which implies
$(1-\epsilon)^{n_o} \lt \frac 1{n_oa} \lt x$
$\lambda=1-\epsilon$ is certainly between $0$ an $1$. The integral
$$
\int_{0}^{1}\lambda^n d \lambda=\frac{1}{n+1}
$$
Converges, hence the integrand converges to $0$. Your result follows.