Prove that $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ without calculus? I'm stuck with this sample RMO question I came across:

Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...

In the solution to this problem, I found the solver making the assumption, 
$\sqrt[n]{n}>\sqrt[n+1]{n+1}$ for $n \geq 3$ How would you prove this?
Any help would be greatly appreciated.
EDIT: In this competition, you aren't allowed to use calculus. Non-calculus methods would be appreciated.

Note that
$$
\begin{align}
\frac{n\left(\frac n{n+1}\right)^n}{(n-1)\left(\frac{n-1}n\right)^{n-1}}
&=\left(\frac{\ n^2}{n^2-1}\right)^{\!\large n}\\
&\ge1\tag1
\end{align}
$$
Inequality $(1)$ says that $n\left(\frac n{n+1}\right)^n$ is increasing. For $n=3$, we get
$$
\begin{align}
n\left(\frac n{n+1}\right)^n
&=3\left(\frac34\right)^3\\
&=\frac{81}{64}\\[6pt]
&\gt1\tag2
\end{align}
$$
Thus, for $n\ge3$, we have $n\left(\frac n{n+1}\right)^n\gt1$. Multiplying by $(n+1)^n$ gives
$$
n^{n+1}\gt(n+1)^n\tag3
$$
and taking the $n(n+1)$ root yields
$$
\sqrt[\large n]{n}\gt\sqrt[\large n+1]{n+1}\tag4
$$
