Proving that the exponential inequality $e^x \ge x^e$ holds for all $x \ge 0$ 
How does one prove that $$e^x \ge x^e$$ for all $x \ge 0$?

I tried to do this by setting $f(x)=e^x-x^e$
Plotting this function shows this easily, as seen here. 
However, when I tried to prove this, it proved quite difficult. It seems to be increasing for $0 \le x \le e$, and seems to be increasing for all $x \ge e$. 
I tried to use that $f'(x)=e^x-ex^{e-1}$ but was not able to. 
Any help would be appreciated. 
 A: Hint Since $x \mapsto \log x$ is a strictly increasing function, applying it to both sides of the inequality gives that it is equivalent to $$x \geq e \log x .$$
There are at least two options here:
(1) Rearrange the inequality as $\frac{\log x}{x} \leq \frac{1}{e}$, and analyze the expression on the l.h.s. of this equality. In particular, show that its maximum value is (at most) $\frac{1}{e}$.
(2) From inspecting the original equation, we know that both sides of the inequality are equal at $x = e$. So, if the inequality holds we must have that the derivatives of both sides of the inequality agree at that point, and indeed, this is straightforward to check. These facts, together with $\frac{d^2}{dx^2} \log x < 0 = \frac{d^2}{dx^2} x$ are sufficient to give the inequality.
A: If $x=0$, then given inequality holds. Suppose that $x > 0$ and define a function $f(x)=x - e\ln x$. The derivative of $f(x)$ is $f'(x)=1-\frac{e}{x}$. Thus $f(x)$ decreases when $x<e$ and increases when $x > e$. $f(e)=e-e\ln e =0$ and $f(e)$ is minimum of $f(x)$, so $f(x)\ge 0$ for all $x > 0$. Thus, $x\ln e \ge e\ln x$ for all $x > 0$ and so $\ln e^x \ge \ln x^e$. Since $\ln x$ strictly increases, $e^x \le x^e$ for all $x > 0$. We previously showed that the inequality holds at $x=0$, so we get the conclusion expected.
A: The graph $\gamma$ of $\log$ is concave, hence $\gamma$ stays below the tangent to $\gamma$ at $(e,\log e)$:
$$\log x\leq \log e+{1\over e}(x-e)\qquad(x>0)\ .$$
This simplifies to $\log x\leq{x\over e}$, and exponentiation gives
$$x^e\leq e^x\qquad(x>0)\ .$$
The case $x=0$ is obvious.
A: Consider the function $$f(x)=e^x-x^e$$ The derivative is $$e(e^{x-1}-x^{e-1})$$
It is clear that $$f'(1)=f'(e)=0$$ and it is verify that $x=1$ goes for a maximun and $x=e$ goes for a minimun. Furthermore $f(1)=e-1>0$ and $f(e)=e^e-e^e=0$.
Taking in account that $f(0)=1$ and the domain of $f$ is $x\ge 0$ it follows the conclusion $$e^x\ge x^e$$ 
