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.

Assume $n\geq 3.$ Start with this calculation:
$$\frac{(n+1)^n}{n^n} = \left(1+\frac{1}{n}\right)^n = 1+\binom{n}{1}\frac{1}{n} +\binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} +\cdots + \frac{1}{n^n}.$$
In the $k$th term, the numerator of the binomial coefficient is $n(n-1)(n-2)\cdots (n-k+1)$ which is less than $n\cdot n\cdots n=n^k.$  So the binomial expansion above is less than 
$$1+1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \frac{1}{n!},$$
which, in turn is less than 
$$1+1+ \frac{1}{2}+\frac{1}{2^2} + \cdots +\frac{1}{2^n} < 3\leq n. $$
So we have $$n > \frac{(n+1)^n}{n^n}$$
$$n^{n+1} > (n+1)^n$$
$$n^{(n+1)/n} > n+1$$
$$n^{1/n} > (n+1)^{1/(n+1)}.$$
