Maximum in given series. Find the maximum term among, $1$, $2^{\frac{1}{2}}$, $3^{\frac{1}{3}}$, $4^{\frac{1}{4}}$, $...$
Now, if we take $f(x) = x^\frac{1}{x}$, and differentiate it is quite simple to see that it reaches it's maximum value at $x=e$ and is increasing on the left of $e$ and decreasing on the right. 
Thus it's obvious that the maxium is either the $\sqrt2$ or $3^\frac{1}{3}$. Now it seems that since $3$ is closer to $e$ than $2$, the greater value should be at $3$ which I confirmed using a calculator, but I can't seem to figure out how to go about proving it on paper, any help?
The curve isn't symmetrical either, so the closer argument isn't very reliable. 

Let $a = 2^{1/2}$ and $b = 3^{1/3}$. Then
$$a^6 = 2^{6/2} = 8$$
while $$b^6 = 3^{6/3} = 9$$
Since $t \mapsto t^6$ is monotone increasing, the result follows.
