Prove for $a > 1$, $n\in \mathbb{N}$, $a^\frac{1}{n} >1$

Question

Prove for any $a > 1$, $n\in \mathbb{N}$, $a^\frac{1}{n} >1$

How do i prove this?

Answer 1

Let $x=a-1>0$ and $n\in \mathbb{N}$, Bernoulli rule shows $$a^\frac{1}{n}=(1+x)^\frac{1}{n}\geq1+\frac{1}{n}x>1$$

Answer 2

It's probably easier to prove that if $0<b\leq 1$, then for each $n\in\mathbb N$, we have $0<b^n\leq 1$. This fact is easy to show by induction, as all you need is that if $0<x,y\leq 1$, then $0<xy\leq 1$.

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Once you have that, you can prove that if $a^{\frac1n}\leq 1$, then $a\leq 1$, which you know because $a=\left(a^\frac1n\right)^n$.

This, along with knowing that $A\implies B$ is the same as $\neg B\implies \neg A$, should be all you need.