Proving $\mathrm e <3$ Well I am just asking myself if there's a more elegant way of proving $$2<\exp(1)=\mathrm e<3$$ than doing it by induction and using the fact of  $\lim\limits_{n\rightarrow\infty}\left(1+\frac1n\right)^n=\mathrm e$, is there one (or some) alternative way(s)? 

First let's consider a simple heuristic argument to show that $2<e<4$.  It is easy to prove using the definition of the derivative that if $f(x)=2^x$ then $f'(x) = (\text{constant}\cdot 2^x$).  The curve gets steeper as $x$ increases, and the average slope between $x=0$ an $x=1$ is $(2^1-2^0)/(1-0)= 1$.  Therefore, the slope at $x=0$ is less than $1$; hence the "constant" is less than $1$.  Now do the same with $g(x)=4^x$ on the interval from $x=-1/2$ and $x=0$, and conclude that the slope at $x=0$ is more than $1$; hence the "constant" you get there is more than $1$.
So $2$ is too small, and $4$ is too big, to serve as the base of the natural exponential function.
It's messier to do the same with $3$, but the interval from $x=-1/6$ to $x=0$ will do it, and you conclude $3$ is too big to be the base of the natural exponential function.
