Help with proof of $(n+1)^n > n! 2^n$

Question

I have already managed to prove it using induction and Bernoulli's inequality but I wonder if there is another way. My proof goes like this: (This is my first time using MathJax, so I apologize for the formatting)
$$(n+1)^n>2^nn!$$ and $$2(n+1)^{n+1}>2^{n+1}(n+1)!$$ for $n>1, n\in \Bbb N$. Using the Bernoulli's inequality that $(1+x)^n>1+nx$ for $n>1$ and $x>-1$, we say $$(1+ \frac 1{n+1})^{n+1}>2$$ $$(n+2)^{n+1}>2(n+1)^{n+1}$$ As $(n+2)^{n+1}>2(n+1)^{n+1}$ and $2(n+1)^{n+1}>2^{n+1}(n+1)!$ then:
$$(n+2)^{n+1}>2^{n+1}(n+1)!$$ We prove with the base case $n=2,$ $(2+1)^2>2^22!$

Answer 1

You may use the AM-GM inequality, for which:

$$ n! = \prod_{k=1}^{n} k \leq \left(\frac{1}{n}\sum_{k=1}^{n}k\right)^n = \left(\frac{n+1}{2}\right)^n, $$ or prove that: $$\forall n>1,\qquad \frac{(n+1)^{n}}{n^{n-1}}>2n $$ that is equivalent to: $$ \forall n>1,\qquad \left(1+\frac{1}{n}\right)^n > 2 $$ that follows from the binomial theorem: $$ \left(1+\frac{1}{n}\right)^n = 1 + \frac{n}{n} + \binom{n}{2}\frac{1}{n^2}+\ldots > 2.$$

Answer 2

Hint:use inequality $\delta _{ n }=\sqrt [ n ]{ { x }_{ 1 }{ x }_{ 2 }...{ x }_{ n } } $ (Geometric mean),${ \xi }_{ n }=\frac { { x }_{ 1 }{ x }_{ 2 }+...+{ x }_{ n } }{ n } $ ( Arithmetic mean)where $${ \xi }_{ n }\ge \delta _{ n }\quad $$, where $x_{ k }=k,\quad k=\overline { 1,n } $

Answer 3

By Stirling's approximation we have $$n!2^{n}<en^{1/2}\left(\frac{2n}{e}\right)^{n}=\left(n+1\right)^{n}\left(\frac{2ne^{1/n}n^{1/\left(2n\right)}}{e\left(n+1\right)}\right)^{n} $$ and $$\left(\frac{2ne^{1/n}n^{1/\left(2n\right)}}{e\left(n+1\right)}\right)^{n}\longrightarrow0 $$ as $n\longrightarrow\infty$.