Show that $x^e\le e^x$ for all $x\gt 0$ and $x \in \mathbb {R}$ I understand that this question may seem quite simple, but although I can see different ways of showing this, I don't understand how it follows from the context I was given (i.e why the second part of the question begins with "hence"). It may also help to bear in mind that I have only just started teaching myself calculus and so far I have only covered differential calculus up to the level taught in secondary schools. 
The question is in two parts (I understand that to answer (i) you simply find $f'(x)$ and evaluate $f'(e)$ to show that it is equal to zero and is therefore, since there is only one stationary point which is a maximum point as implied by the question, the maximum point): 
$f(x) = {\ln x\over x}$, $x\gt 0$
(i) Show that the maximum point on the graph of $y = f(x)$ occurs at the point $\left(e,\frac{1}{e}\right)$. 
(ii) Hence, show that  $x^e\le e^x$ for all $x\gt 0$
Any help would be greatly appreciated.

Consider the function $F(x)= x^{\frac 1x}$.  We claim that $F(x)$ has a maximum at $x=e$.  Indeed your $f(x)$ is $\log F(x)$ so this follows from what you have proved.  But this then means that $$x^{\frac 1x}\leq e^{\frac 1e}\implies x^e\leq e^x$$
