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

This is the same as comparing $\frac{3}{2}\log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $\frac{1}{4}$ on $(0,1)$, we have $$ 0\leq\int_{0}^{1}\frac{x^2(1-x)^2}{1+x}\,dx \leq \frac{1}{16}$$ where the middle integral is exactly $-\frac{11}{4}+4\log(2)$. It follows that $$ \frac{33}{32} \leq \frac{3}{2}\log(2) \leq \frac{135}{128} $$ so $\frac{3}{2}\log(2)>1$ and $\color{red}{2\sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.

About $\sqrt{2}\log(2)$, we have $$ \log(2)=\lim_{n\to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}\leq\lim_{n\to+\infty}\sum_{k=n+1}^{2n}\frac{1}{\sqrt{k}\sqrt{k-1}}\stackrel{\text{CS}}{\leq}\lim_{n\to +\infty}\sqrt{n\sum_{k=n+1}^{2n}\left(\frac{1}{k-1}-\frac{1}{k}\right)}$$ and the RHS is exactly $\frac{1}{\sqrt{2}}$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

Instead of comparing $2^{\sqrt 2}$ and $e$, let's raise both to $\sqrt 2$ and compare $2^2$ and $e^{\sqrt 2}$: $$ e^{\sqrt 2} > 2.7^{1.4} \approx 4.017068799 > 4 = 2^2 $$ Or use that $$ e^x > 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} $$ with $x=1.41$ and get $$ e^{\sqrt 2} > e^{1.41} > 4.03594 > 4 $$ In fact, $e^{\sqrt 2} \approx 4.113250377 > 4$.

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 $2$:

$$\ln(2)=\int_1^2{dx\over x}\lt{1\over6}\left(1+2\cdot{3\over4}+2\cdot{3\over5}+{1\over2} \right)={1\over6}\left(1+{3\over2}+{6\over5}+{1\over2} \right)={1\over6}\cdot{42\over10}={7\over10}$$