# How to prove that $n^{1/n}>(n+1)^{1/(n+1)}$ is equivalent to $n^{n+1}>(n+1)^n$?

Found this inside an example in my analysis book. Can't figure out why the two are equivalent and how to work out the algebra.

My question is:

• What is the algebra behind $n^{1/n}>(n+1)^{1/(n+1)}$ is equivalent to $n^{n+1}>(n+1)^n$ (and how to prove it)?

Sorry if this is something basic, I'm just not being able to work out the algebra behind it.

EDIT: I think I need to take a break...

$$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ it's $$\left(n^{\frac{1}{n}}\right)^{n(n+1)}>\left((n+1)^{\frac{1}{n+1}}\right)^{n(n+1)}$$ or $$n^{n+1}>(n+1)^{n}$$
$f(x)=x^{n(n+1)}$ is strictly increasing for $x\geq 0$.
• $f(x)=x^6$ is strictly increasing ? – Aqua Jan 7 '18 at 17:48
• It means $x\geq0$. – Michael Rozenberg Jan 7 '18 at 17:50
powering by $n(n+1)$ we get $$n^{n+1}>(n+1)^n$$ and it follows that $$\left(1+\frac{1}{n}\right)^n<n$$ for $n\geq 2$