# How can one prove that e<πe<\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:

1. Prove (without calculator) that $e<\pi$;

2. 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.