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.
 A: consider the function $f(x)=x^{1/x}$ for $x >0$. Now check it's monotonicity.
\begin{align*}
\ln f(x) & = \frac{\ln x}{x}\\
\frac{f'(x)}{f(x)} & = \frac{1-\ln x}{x^2}\\
f'(x) & = f(x)\left[\frac{1-\ln x}{x^2}\right]
\end{align*}
Observe that for $x >e$, we get $f'<0$. So for $n \geq 3$, we have $f(n) > f(n+1)$.
A: 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 $$n > \frac{(n+1)^n}{n^n}$$
$$n^{n+1} > (n+1)^n$$
$$n^{(n+1)/n} > n+1$$
$$n^{1/n} > (n+1)^{1/(n+1)}.$$
A: We need to prove that $$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$ or
$$\frac{\ln{n}}{n}>\frac{\ln(n+1)}{n+1}.$$
Let $f(x)=\frac{\ln{x}}{x}$, where $x>0$.
Thus, $$f'(x)=\frac{\frac{1}{x}\cdot x-\ln{x}}{x^2}=\frac{1-\ln{x}}{x^2}<0$$
for all $x>e$.
Thus, for all $n\geq3$ we have
$$n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}.$$ 
Now, for $n=2$ we get
$$\sqrt2<\sqrt[3]3,$$
and for $n=1$ we have $1^1<2^\frac{1}{2}$, which gives that $\sqrt[3]3$ is a largest number in the  sequence.
Done!
A: HINT.-The function $f(x)=x^{\frac 1x}$ has as derivative $f'(x)=-x^{\frac 1x-2}(\ln(x)-1)$ which prove that $f$ is decreasing on $x\gt e$
You can deduce that the maximum value is $\color{red}{\sqrt[3]3\approx1.442249}$ (verifying that $\sqrt2\approx1.414243$).
A: 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 $n\cdot n^n$.
A: 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
$$
\sqrt[\large n]{n}\gt\sqrt[\large n+1]{n+1}\tag4
$$
