Quick but not simple question. $2^\sqrt2$ or e, which is greater? 
$2^\sqrt2$ vs $e$, which is greater?

$(2^\sqrt2)^\sqrt2 = 4\quad $  &  $\quad e^\sqrt2$ = ?
$\log(2^\sqrt2) = \sqrt2\log(2)\quad$ & $\quad \log(e) = 1$
I tried but can't induce comparable form.
Is anybody know how to prove it?

If you know that $\ln(2)\approx0.69$ and $1/\sqrt2=\sqrt2/2\approx1.414/2=0.707$, then you have $\ln(2)\lt1/\sqrt2$, in which case $\ln(2^\sqrt2)=\sqrt2\ln2\lt1=\ln(e)$, hence $2^\sqrt2\lt e$.
It's not hard to show that $\sqrt2\gt1.4$, since $1.4^2=1.96\lt2$.  It's a little trickier to show that $\ln(2)\lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:
$$\ln(2)=\int_1^2{dx\over x}\lt{1\over6}\left(1+2\cdot{3\over4}+2\cdot{3\over5}+{1\over2} \right)={1\over6}\left(1+{3\over2}+{6\over5}+{1\over2} \right)={1\over6}\cdot{42\over10}={7\over10}$$
