I was perusing the wikipedia page on Mathematical induction, and it mentions it's possible to prove by induction that.
$\frac{n^{n}}{3^{n}}<n!<\frac{n^{n}}{2^{n}}$ for $n\geq6$
Proof for $n=6$ is obvious.
I tried to complete the proof by trying to prove that:
$$\frac{\left(n+1\right)^{\left(n+1\right)}}{3^{\left(n+1\right)}}<\left(n+1\right)!<\frac{\left(n+1\right)^{\left(n+1\right)}}{2^{\left(n+1\right)}}$$ Expanding the terms: $$\frac{\left(n+1\right)^{n}\left(n+1\right)}{3^{n}\cdot3}<n!\left(n+1\right)<\frac{\left(n+1\right)^{n}\left(n+1\right)}{2^{n}\cdot2}$$ Cancelling: $$\frac{\left(n+1\right)^{n}}{3^{n}\cdot3}<n!<\frac{\left(n+1\right)^{n}}{2^{n}\cdot2}$$ Which together with the initial inequalities yields: $$2\leq\frac{\left(n+1\right)^n}{n^n}\leq3$$ Which is where I'm stuck. I can see that the middle term of the inequality tends to e and that it holds up, but I don't see how I can prove it. I thought maybe something with logaritms, but I can't quite get it.
Or am I barking up the wrong singly connected undirected graph?
Expanding and keeping only the first two terms, $$(n+1)^n=n^n+n\ n^{n-1}+\frac12n(n-1)\ n^{n-2}+\frac1{3!}n(n-1)(n-2)\ n^{n-3}...>2\ n^n.$$
Expanding, using $n-k<n$ and $k!>2^k$, $$(n+1)^n<n^n+n\ n^{n-1}+\frac12n^2\ n^{n-2}+\frac1{2^2}n^3\ n^{n-3}...=3\ n^n.$$ In passing, this proves that $2<e<3$.