easy  general 
Without explicitly calculating, find out which is bigger: $e^\pi$ or $\pi^e$?
One method would be to expand $e^x$, another would be to observe the derivative of a special function
$e^\pi$
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.
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.
Notes:

This puzzle is kept in the "easy" category because it is easy to ask. The answer is also "easy" to verify, using a calculator. But when solving, it is a hard puzzle.

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

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 $x=\sqrt{e}, \sqrt{\pi}$