Which one is greater?

$\because e^{x}=1+x+\frac{x^{2}}{2!}+...$ $\therefore e^{x}> 1+x.$ Substitute in $x=\frac{\pi }{e}-1$ , giving $e^{\frac{\pi}{e}-1}> \frac{\pi }{e}$ $\Rightarrow e^{\frac{\pi}{e}}> \pi$ $\Rightarrow e^{\pi}> \pi ^{e}.$

Consider a function f(x) = $x^{ \dfrac{1}{x} }$

$$ Differentiating with respect to x,

$$ f '(x) = $x^{ \dfrac{1}{x} } \times [ \dfrac{1}{x^2} - \dfrac{ln(x)}{x^2} ]$

$$ When f(x) is decreasing f '(x) < 0,

$$

Since $x^{ \dfrac{1}{x} }$ & $\dfrac{1}{x^2}$ are greater than 0, it implies $1 - ln(x) < 0$

$$ $ln(x) > 1$

$$ $x > e$

$$ Hence, f(x) is decreasing in the region, x > e. $$

By definition of decreasing function as,

$\pi > e$

$$ $f( \pi ) < f(e)$

$$ $\pi^\dfrac{1}{\pi} < e^\dfrac{1}{e}$

$$ Raising to $\pi \times e$ on both sides,

$$ $\boxed{ \pi^{e} < e^{\pi}}$

$\begin{aligned} e^\pi &\le=\ge \pi^e \\ \pi \ln e &\le=\ge e \ln \pi \\ \ln e \over e &\le=\ge {\ln \pi \over \pi}\\ \text{So we look at the function } \\ f(x) &= {\ln x \over x}\\ \text{which reaches the maxima at} \\ \frac{{1 \over x} x -1 \ln x}{x^2} &= 0\\ \text{or, } x&=e \\ \text{Hence, }{\ln e \over e} &\ge {\ln \pi \over \pi} \\ e^\pi &\ge \pi^e\\ \end{aligned}$

Use Taylor expaniaon

It is true that e^x> x^e if x doesn't equal e

I did it like, 3^4 is greater than 4^3 As π is greater than e, thus e^π > π^e xD

e to the power of any real number other than e itself will be greater than the number to the power of e

Just look at the exponent; pi = 3.14 but Euler's = 2.71. Therefore, 2.71^3.14 > 3.14^2.71