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.

We wish to compare $\sqrt[n]n \lessgtr \sqrt[n+1]{n+1}$. Raise each side to the $n(n+1)$th power to get
$$ n^{n+1} \lessgtr (n+1)^n $$
and use the binomial theorem on the right-hand side:
$$ n\cdot n^n \lessgtr \underbrace{n^n+\binom n1 n^{n-1} + \binom n2 n^{n-2} + \cdots + \binom n{n-1} n^1}_{n\text{ terms}} + 1 $$
Because $\binom{n}{k}\le n^k$, each of the $n$ indicated terms is at most $n^n$. And when $n\ge 3$, the last term $\binom n{n-1}n^1 = n^2$ is so much smaller than $n^n$ that the final $1$ term is insufficient to make the RHS exceed $n\cdot n^n$.
