Nth root of n is greater than 1?

Question

A proof I did recently called upon a "fact" which my prof called without giving explanation or proof, which is the "fact" that $\sqrt[n]{n}>1$, how can this be shown?

Answer 1

Suppose $x>0$.

If $p>0$, then $x^p>1 \iff x>1$, and $x^p<1 \iff x<1$.

If $p<0$, then $x^p<1 \iff x>1$, and $x^p>1 \iff x<1$.

The second statement follows from the first by just taking reciprocals, which reverses the inequality.

The first statement is true because the function $f(p)= x^p$ is a strictly increasing function of $p$ for fixed $x>1$, and a strictly decreasing function of $p$ for fixed $x<1$.

In your case, take $p=1/x$ (a positive number) and you are considering $x=n$ for integral $n>1$ (so certainly $x>0$).

Answer 2

Let $x = \sqrt[n]{n} \quad for \quad n>1$

$x^n=n$

Assume $0<x\leqslant1$, then $x^n\leqslant x^{n-1}\leqslant...\leqslant1 <n$

There is a contradiction, so $x>1$

Answer 3

It follows trivially (raise both sides to the power of $\frac{1}{n}$), however $n$ must be greater than $1$. Take a look at a graph of $x^{1/x}$: it is less than $1$ for all $x<1$.