Compare $e^2$ and $7$ without using calculator Which is bigger? $e^2$ or $7$? Any tricks?
Don't know quite how to approach those kind of things. 
 A: If you know that $e > 2.7$, then $e^2 > 2.7^2 = \dfrac{27^2}{10^2} = \dfrac{729}{100} = 7.29 > 7$.
A: $$e=\sum_{k=0}^{\infty}\frac{1}{k!}\gt \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=\frac 83>2.66\Rightarrow e^2\gt (2.66)^2=7.0756\gt 7$$

As Mark Bennet pointed, the following is much easier : 
$$e=\sum_{k=0}^{\infty}\frac{1}{k!}\gt \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=\frac 83\Rightarrow e^2\gt \left(\frac 83\right)^2=\frac{64}{9}\gt 7$$
A: Here is a hint: $e^x=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\dots$
This is a rapidly convergent series, so either you should be able to exceed a bound like $7$ with a few terms, or see that it will  be less, and use a comparison (e.g. with a geometric progression) to prove that it remains less.
Try $x=2$ for size (easy enough to do by hand without a calculator).

For the calculation, since others have put theirs:

$$1+2+\frac 42+\frac 86+\frac {16}{24}=7$$ and there are other positive terms.

A: If one recalls that $e>\frac{19}{7}$ (from the continued fraction representation of $e$) then it is straightforward to check that:
$$ \color{red}{e^2} \color{purple}{>} \frac{361}{49} = \color{red}{7}+\frac{18}{49},$$
i.e. one just needs to check that $19^2>7^3$.
A: Here's an alternative. Try to compare $2$ and $\ln(7)$. $\ln(x)$ has a more natural geometric interpretation as area under $y=\frac{1}{x}$, compared to $e^x$, so some geometry might avail us. 
Since $\ln(7)=\int_1^7\frac{1}{x}\,dx$, we consider approximating the integral with a trapezoid rule style Riemann sum. The trapezoids closer to $1$ will do a worse job since concavity is higher there, so we want these trapezoids to have a narrower base. Using a linear progression of base width would mean we need to cut up $[1,7]$ into $T_n$ equal sized pieces, where $T_n$ is a triangular number. Using $T_n=15$ (and therefore $5$ trapezoids with bases $\frac{2}{5}$, $\frac{4}{5}$, $\frac{6}{5}$, $\frac{8}{5}$, $\frac{10}{5}$) gives us:

And we can see/calculate that $$\ln(7)<\frac{1}{2}\left(\frac{2}{5}\left(1+\frac{5}{7}\right)+\frac{4}{5}\left(\frac{5}{7}+\frac{5}{11}\right)+\frac{6}{5}\left(\frac{5}{11}+\frac{5}{17}\right)+\frac{8}{5}\left(\frac{5}{17}+\frac{1}{5}\right)+\frac{10}{5}\left(\frac{1}{5}+\frac{1}{7}\right)\right)$$
$$\ln(7)<\frac{65376}{32725}<2$$
So $7<e^2$.
