Initial Misstep

$e^\pi \approx 23.14; \pi^e \approx 22.46$, hence $e^\pi$ is bigger. But this is not a valid method. We need to compare the two numbers without calculating them.

Method 1

Since $e^x = 1 + x + \dfrac{x^2}{2!} + \ldots$, we have $e^x > 1+x$.

Let $x = (\dfrac{\pi}{e} -1)$ and solve:

$\implies e^{\pi/e - 1} > 1 + \dfrac{\pi}{e} - 1$

$\implies \dfrac{e^{\pi/e}}{e} > \dfrac{\pi}{e}$

$\implies e^{\pi/e}$ > $\pi$

$\implies e^\pi$ > $\pi^e$ (raised to power $e$ on both sides)

Hence $e^\pi$ is bigger.

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Method 2

Suppose $\sim$ denotes the relation between the two terms.

$e^\pi \sim \pi ^ e \iff e^{1/e} \sim \pi^{1/\pi}$ (raised to power $1/{e\pi}$ on both sides)

Note the function $f(x) = x^{1/x}$

To investigate how this function behaves, we need to take its derivative. But that's difficult.

Consider $g(x) = \ln(f(x)) = \dfrac{\ln(x)}{x}$

Differentiate using product rule:

$(u \cdot v)' = u' \cdot v + u \cdot v'$

where $u=\ln(x), v = 1/x, u'=1/x, v' = -1/x^2$

$g'(x) = \dfrac{1}{x} \cdot \dfrac{1}{x} + \ln(x) \cdot \dfrac{-1}{x^2}$

$g'(x) = \dfrac{1-\ln{(x)}}{x^2}$

The derivative $g'$ is negative if $\ln(x) > 1$

$\implies g$ is decreasing for $x > e$

$\implies f$ is decreasing for $x > e$ because $\ln$ is a monotonic transformation.

$\implies f(e) > f(\pi)$ because $e<\pi$

$\implies e^{1/e} > \pi^{1/\pi}$

$\implies e^\pi > \pi^e$

Thus $e^\pi$ is bigger.

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Notes:

1. This puzzle is usually considered easy because it is easy to ask and verify (using a calculator). But when solving, it is a more difficult than many of the hard puzzles here.

2. Here's a plot of $f(x)=x^{1/x}$

   

3. This is a common question in placement tests. It would be a good idea to memorize that $f(x) = x^{1/x}$ is decreasing in the range $x > e$. Using this, we can prove other variants.

   	relation			using function
   $e^{\pi}$	$>$	$\pi^{e}$		$f(x)$ at $x=e, \pi$
   $e^{2 \pi}$	$>$	$\pi ^ {2e}$		$f^2(x)$ at $x=e, \pi$
   $e^{\pi^2}$	$>$	$\pi ^ {e^2}$		$\sqrt{f(x)}$ at $x=e^2, \pi^2$
   $e^{\sqrt{\pi}}$	$<$	$\pi ^ {\sqrt{e}}$		$f(x)^2$ at $x=\sqrt{e}, \sqrt{\pi}$