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.

By i) you have determined
$f(x) = \frac {\ln x}x \le \frac 1e; \forall x > 0$.
So $e \ln x \le x$
So $e^{e\ln x} \le e^x$
And  $e^{e\ln x}=(e^{\ln x})^e = x^e \le e^x$.
==
Part i) is a matter of taking $f'(x)$ and setting to zero.
$f'(x) = (\frac {\ln x}x)' = \ln x(-\frac 2{x^2}) + \frac 1x*\frac 1x  = \frac{1- \ln x}{x^2}$
$f'(x) = 0 \implies$
$\ln x = 1$ 
$x = e$
$f''(x) = -\frac 1{x^2} - 2\ln x \frac 1{x^3} < 0$ for $x = e$ so $x=e; f(x) = \frac 1e$ is maximum and for all $x > 0; \frac {\ln x}x \le \frac 1e$.
