Compare which of $n!$ and $\left(\dfrac n3\right)^n$ will stay larger for $n$ large enough?
The answer is undoubtedly $n!>\left(\dfrac n3\right)^n$ and I'm trying to prove this.
My first approach was to prove that $$n+1>\dfrac{(n+1)^{n+1}}{3n^n}\tag{1}$$ for $n$ large enough. This could establish the inductive relationship $$n!>\left(\dfrac n3\right)^n\implies(n+1)!>\left(\dfrac{n+1}3\right)^{n+1}$$thus solve the problem. (Geogebra graph told me this is true) However I found $(1)$ hard to prove.
My second approach was the Stirling formula$$n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n.$$So$$n!>\left(\dfrac n3\right)^n\Leftrightarrow\sqrt[2n~~~]{2\pi n}>\frac e3.$$ I again failed to complete the proof. I'm now out of ideas so I need help.
Rearranging (1) yields $$3 > \left(1 + \frac{1}{n}\right)^n.$$ The right-hand side converges to $e$ as $n \to \infty$.