TITLE: Is My Proof that $\pi^e < e^{\pi}$ Valid?
QUESTION [5 upvotes]: The other day, a math teacher at my college gave me a challenge problem: Prove that $$\pi^e < e^{\pi}$$ without using a calculator. The next day, I found a valid proof, but I used a log table instead of a calculator, so my proof is hardly satisfying. I went looking for another proof and I came up with something that might not be legitimate. Here's my proof:
Suppose that is an $x$ such that $x^e < e^x$
Then taking the natural log of both sides,
$$\ln(x^e) < \ln(e^x)$$
$$e\ln(x) < x$$
$$\ln(x) < \frac{x}{e}$$
Differentiate both sides,
$$\frac{1}{x} < \frac{1}{e}$$
$$x>e$$
Thus, $x^e < x^e$, for $x > e$
Since $\pi > e$, therefore $\pi^e < e^{\pi}$
I think my proof is good except perhaps where I differentiated both sides of the inequality. I know that there are many cases where this is invalid, but I'm not sure about this case.

REPLY [7 votes]: As others have pointed out, you cannot differentiate and think the inequality remains valid. However, you can integrate, retracing your steps. 
$$x>e\implies \frac1x<\frac1e\implies \int_e^{\pi}\frac1x\,dx<\int_e^{\pi}\frac1e\,dx\implies \log \pi < \frac{\pi}e\implies \pi^e<e^\pi$$