Which number is greater, $2^\sqrt2$ or $e$?

To show $2^{\sqrt{2}}<e$, it suffices to show that $\sqrt{2}\ln 2 < 1$, or $\ln 2 < \frac{1}{\sqrt{2}}$. Notice that $\ln 2 = \int\limits_{1}^{2}{\frac{1}{x}\,dx}$, and by Cauchy-Schwarz we have $$\ln 2 = \int\limits_{1}^{2}{\frac{1}{x}\cdot 1\,dx}\le\left(\int\limits_{1}^{2}{\frac{1}{x^2}\,dx}\right)^{1/2}\left(\int\limits_{1}^{2}{1\,dx}\right)^{1/2} = \left(\frac{1}{2}\right)^{1/2}\left(1\right)^{1/2} = \frac{1}{\sqrt{2}}.$$ Equality cannot hold as $(1/x)/1$ is not constant, so it follows that $\ln 2 < \frac{1}{\sqrt{2}}$, as desired.

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 $2\cdot 5^2>7^2$).

Approximating the integral by the area of a trapezoid we get for $x>1$ $$\log(x)=\int_1^x\frac{\mathrm{d}t}{t}<(x-1)\frac{1+\frac1x}2=\frac{x^2-1}{2x}$$ So $$\sqrt{2}\log(2)=2\sqrt{2}\log(\sqrt{2})<1.$$