Which number is greater, $2^\sqrt2$ or $e$? 
Claim: $\color{red}{2^\sqrt2<e}$

Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus 
$$ f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln  x \implies e^{x-1} > x^{\sqrt{x}}  $$
Putting $x=2$, we obtain
$$e>2^\sqrt2$$
What do you think about this approach? Is there an easier way?

The inequality $e>2^\sqrt2$ is equivalent to 
$$e^{\sqrt{2}}>(2^\sqrt2)^{\sqrt{2}}=2^2=4.$$
Now
$$e^{\sqrt{2}}>\sum_{k=0}^5\frac{(\sqrt{2})^k}{k!}=\frac{13}{6}+\frac{41\sqrt{2}}{30}>\frac{13}{6}+\frac{4}{3}\cdot\frac{7}{5}=\frac{121}{30}>4$$
because $\sqrt{2}>7/5$ (equivalent to $2\cdot 5^2>7^2$).
