# Prove that for $x>1$ there exists $n\in\mathbb N\colon y < x^n$.

Let $x,\,y \in \mathbb{R}$ such that $x > 1, y > 0$. Using reductio ad absurdum prove $$\exists n \in \mathbb{N}\colon y < x^n$$ Suggestion: if $\varepsilon > 0$ is small and $n\in \mathbb{N}$, then there exists $n$ for which $x^n + \varepsilon < x^{n+1}$.

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You may want to provide your own thoughts on the problem before anyone comments on it. –  gt6989b Nov 14 '12 at 23:01

Hint: Let $x=1+t$. By the Bernoulli Inequality (easily proved by induction on $n$), we have $(1+t)^n \ge 1+nt$.

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Let's assume contrary, i.e. that for any $n \in \mathbb N$ $y>x^n$.

Let's take $\varepsilon = x - 1 > 0$. Let $m \in \mathbb N$ be the smallest number for which $m\varepsilon \ge y$ i.e. $m=\lceil{\frac{y}{\varepsilon}}\rceil$.

we have by assumption: $y > x^m =(1+\varepsilon)^m$

which, as a power of a binomial can be extended as $(1+\varepsilon)^m =1 + m\varepsilon + \ldots + m\varepsilon^{m-1} + \epsilon^m > 1 + m\varepsilon > y$ thus we get contradiction since $y \not> y$

$↯$

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Wlog you can suppose that $y > 1$ since otherwise $y \leq 1 < x$.
Your statement is then equivalent to $\forall y > 1, \forall x > 1, \exists n \in \mathbb{N}: \ln y < n \times \ln x$, which is true since $\mathbb{R}$ is Archimedean.