How can one prove that $e<\pi$? This question is inspired by another one, asking to prove that something approximately equal to $1.2$ is bigger than something approximately equal to $0.9$. The numerical answer to this question was (expectedly) downvoted, though in my opinion it is the most reasonable approach to this kind of problems (${\tiny \text{which I personally find completely useless}}$).

My question will consist of 2 parts:


  
  Prove (without calculator) that $e&lt;\pi$;
  Explain what do we learn from the proof/what makes this problem interesting.
  


Edit: Existing answers only confirm my point of view about various weird inequalitites. Fortunately there is $3$ between $e$ and $\pi$, otherwise the things would be very boring.
Inscribe a regular hexagon in a circle of radius $1$. Since a straight line is the shortest distance between two points the circumference of the circle is longer than the circumference of the hexagon. We take the definition of $\pi$ as half the circumference of the unit circle.

Putting all this together we obtain $2\pi \gt 6$ or $\pi \gt 3$

We take $e$ as the sum $1+1+\frac 12+\frac 1{3!}+\cdots$ which converges absolutely and which, after the first three terms, is term by term less than the sum $1+1+\frac 12+\frac 1{2^2}+\cdots$ since the later terms in the second sum are obtained by dividing the previous term by $2$, and in the first sum by $n\gt 2$ (crudely for $n\ge 3$ we have $n!\gt 2^{n-1}$). 

Summing the geometric series we have $e\lt 3 \lt\pi$.

What do we learn - well how easy it is to make an estimate depends on the definition. The geometric definition of $\pi$ lends itself to a good enough estimate. There are different ways of defining $e$ too, but the sum offers a range of possibilities for estimating, particularly as the terms decrease very quickly. But the geometric definition for $\pi$ requires assumed knowledge about a straight line as the shortest distance between two points, which seems obvious - yet conceals the trickiness of defining the length of a curve - so this looks simpler than it is.
Use the Leibniz formula for $\pi$:  $$
\frac{\pi}{4} = \sum_{j=0}^\infty \frac{(-1)^j}{2j+1}  \text{.}
$$  This is an alternating sum where each partial sum is alternately an upper and lower bound.  The first few terms constrain $\pi$ to these ranges:

[0,4]
[2.666..., 4]
[2.666..., 3.4666...]
[2.895..., 3.4666...]
[2.895..., 3.33968...]
[2.9760..., 3.33968...]
[2.9760..., 3.28374...]
[3.01707..., 3.28374...]


at which point we know $\pi$ is greater than $3$.

A similar method is this descending series for $\mathrm{e}$:  $$
\mathrm{e} = 3 + \sum_{k=2}^\infty \frac{-1}{k!(k-1)k}  \text{.}
$$
Now after one term, we know $\mathrm{e} &lt; 3-\frac{1}{4} &lt; 3$ and the desired inequality follows.

For determining which of two unknown numbers given only by series is larger, it is helpful to have a series whose partial sums are bounds of some sort.  Since we know we want $\mathrm{e} &lt; \pi$ it is helpful to have a decreasing series for $\mathrm{e}$, an increasing series for $\pi$, and/or alternating series for either or both.

Further lesson: It is very nice to have intervals (or at least bounds) on unfamiliar numbers.  Alternating series are quite handy for this.

Other series for this method: \begin{align}
\mathrm{e}^{-1} &amp;= \sum_{k=0}^\infty \frac{(-1)^k}{k!} &amp;&amp;\text{gives $\mathrm{e}&lt;3$ after $5$ terms} \\
\frac{\pi^2}{8} &amp;= \sum_{k=1}^\infty \frac{1}{(2k-1)^2}  &amp;&amp;\text{gives $3 &lt; \pi$ after 4 terms}
\end{align}

Further further lesson: the more representations of a thing you know, the more you are able to show about it.