Show that $\lim_{n \to \infty} \frac{1}{n!} \frac{n^{n+1}}{e^n} = \infty$ Show that 
$$\lim_{n \to \infty} \frac{1}{n!} \frac{n^{n+1}}{e^n} = \infty$$
Wihout using stirlings aproximation to n!
I've tried to compare this to a divergent sequence but didnt work. Also, I dont see how to use L'Hospital or slmething like this.
Thanks in advance!
 A: Lemmas:


*

*This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$

*You can try to prove these inequalities yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k}  \right)^{k+{1 \over 2}}\\$ . We will only need the second inequality, but both are easily provable by taking the logarithm of both sides.
Now,

  
*
  
*We first write $n^n/n!$ in a better way: $\displaystyle \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n \cdot \prod_{i=1}^{n-1}\prod_{k=1}^{i}  \left(1+\frac{1}{k} \right)^{-1}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n\cdot \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{-(n-i)}=\prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i}$
  
*Then, a lower bound: $\displaystyle \prod_{i=1}^{n-1}  \left(1+\frac{1}{i} \right)^{i} \geq \prod_{i=1}^{n-1}\left (e^{1} \cdot  \left(1+\frac{1}{i}\right)^{-1/2} \right )=\frac{e^{n-1}}{\sqrt[2]{n}}$
  
*Finally, $\displaystyle {n^n \over e^n n!}\geq n^{-1/2}e^{-1} \Rightarrow {n^{n+1} \over e^n n!}\geq {n^{1/2} \over e}$ and we are done.
  

A: Maybe this observation could help as a start $$\frac{u_{n+1}}{u_n}=  \frac{\left(1+\frac1n\right)^{n+1}}{e}>1$$  with $$u_n =\frac{1}{n!} \frac{n^{n+1}}{e^n}$$ 
A: By Stirling, $\frac{1}{n!}\bigl(\frac{n}{e}\bigr)^{n-1} \simeq \frac{e}{\sqrt{2\pi}n^{3/2}} \to 0$ if $n \to \infty$. Thus, 
$$
\lim_{n \to \infty}\frac{n^{n+1}}{n!e^n} = \infty
$$ 
A: $$ a_n = \frac{n^n}{n!\,e^n} < \frac{n^{n+1}}{n! \,\,e^n}$$
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1} \, e^n \, n!}{e^{n+1} (n+1)! n^n} = (1 + \frac 1n)^n \to  e > 2 $$ 
$$ \exists\, n_0: \,\, \forall n > n_0 \,\, a_{n+1} > 2 a_n $$
$$  a_{n_0+p} > 2^p a_{n_0}\rightarrow \lim_{n \to \infty} a_n = \infty \rightarrow \lim_{n \to \infty}\frac{n^{n+1}}{n!\,\, e^n} = \infty $$
