Solving this inequality: $e^x \geq x^e$

Show that $e^x \geq x^e$ for $ 0 \lt x \lt \infty $.

I tried to apply the normal logarithm here, which yields $x \geq e\times \ln(x)$ Still, I am kind of stuck here, anyone mind giving me a hand?

Hint

Consider the function $f(x) = e^x - x^e$ and find its minimum over positive $x$. If you do it correctly (e.g. solve $f'(x) = 0$, etc) you will find that it is non-negative. Hence, the claim will follow.

UPDATE

Easier to take logs: $e^x \ge x^e$ iff $x \ge e \ln x$, and now look at $f(x) = x - e \ln x$, which should be elementary

Hint: $e^{x/e}$ is convex and $y=x$ is its tangent at $x=e$.