Largest element in the set $\{1,\sqrt{2},\sqrt[3]{3},...,\sqrt[n]{n}\}$ What is the largest element in the set $\;\left\{1,\sqrt{2},\sqrt[3]{3},...,\sqrt[n]{n}\right\}$?
Once I write down the numbers, it seems like the largest element will be $\sqrt[3]{3}$, but I couldn't come up with an explicit proof. 
It looks like the sequence $a_n=\sqrt[n]{n}$ increases up to $n=3$ and then decreases down to $1.$ 
 A: It's possible to answer this without calculus as well. Consider two consecutive terms $\sqrt[n]n$ and $\sqrt[n+1]{n+1}$, and compare them:
$$
\sqrt[n]n\,?\,\sqrt[n+1]{n+1}\\
n^{n+1}\,?\,(n+1)^n\\
n\,?\,\left(\frac{n+1}{n}\right)^n\\
n\,?\,\left(1+\frac1n\right)^n
$$
where the right-hand side is well-known to never exceed $e$ (and even without that, showing that it never exceeds $3$ is not too difficult). So clearly, for $n\geq3$, the question mark will be a $\geq$, meaning all terms after $\sqrt[3]3$ will be smaller than $\sqrt[3]3$.
A: By the comment and the answer hint which was given I managed to setup the solution. 
Observe that $$f'(x)=\frac{f(x)(1-\ln(x)}{x^2}$$   So the function increases upto $e$ and then decreases from there onwards.
And we can notice that $f(2)=f(4)$ and also both 3 and 4 lie on the decreasing side. Thus $f(3)>f(4)$.
Hence $f(3)$ should be the maximum value. 
A: The function $f(x)=x^{1/x}$ is increasing wherever $g(x)=\log f(x)$ is increasing. Now
$$
g(x)=\frac{\log x}{x}
$$
and
$$
g'(x)=\frac{1-\log x}{x^2}
$$
Can you finish?
