How do you prove that $n^{1/n}>1$ for all $n\geq2$? Induction seems like a good approach. Let's try it.
Base Case: When $n=2$, $n^{1/n} = 2^{1/2}=\sqrt{2}>1$
Induction Step: Let $k\in \mathbb{N}$ and suppose $k^{1/k} > 1$. Then $(k+1)^{1/(k+1)}...$
I am not sure how to proceed from here. Obviously we must use the Induction Hypothesis at some point, but how do you separate $k^{1/k}$ from $(k+1)^{1/(k+1)}$? Or is there a better approach than induction?
 A: $x^{1/n}$ is an increasing function, so if $n>1$, then $n^{1/n}>1^{1/n}=1$
A: $HINT$
$$n^{\frac{1}{n}}=e^{\frac{\log{n}}{n}}$$
Do induction now and use the fact that $\frac{\log{n+1}}{n+1} \geq  \frac{\log{n}}{n+1}=\frac{\log{n}}{n}\frac{n}{n+1}$
A: For sure, induction seems to be the simplest way to do it.
You could also do it using Taylor series. Let $n=p+1$
$$y=n^{\frac{1}{n}}=(p+1)^{\frac{1}{p+1}}\implies \log(y)=\frac{1}{p+1}\log(p+1)=\frac{1}{p+1}\left(\log(p)+\log \left(1+\frac{1}{p}\right) \right)$$ Expanding for large value of $p$
$$\log(y)=\frac{\log (p)}{p}+\frac{1-\log (p)}{p^2}+O\left(\frac{1}{p^3}\right)$$
$$y=e^{\log(p)}=1+\frac{\log (p)}{p}+\frac{\log ^2(p)+2 (1-\log (p))}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ that is to say
$$y=e^{\log(p)}=1+\frac{\log (p)}{p}+\frac{(\log (p)-1)^2+1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$
A: A very elementary way is using the binomial formula


*

*$a^n-1 = a^n-1^n = (a-1)(a^{n-1} + \cdots + a + 1)$ for $n \in \mathbb{N}$, $n \geq 2$.


From this it follows immediately for $a>0$:


*

*$a^n > 1 \Leftrightarrow a > 1$
Now, apply this to $a = \sqrt[n]{n}$ for $n \in \mathbb{N}$, $n \geq 2$.
