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!
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.
$$ a_n = \frac{n^n}{n!\,e^n} &lt; \frac{n^{n+1}}{n! \,\,e^n}$$
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1} \, e^n \, n!}{e^{n+1} (n+1)! n^n} = (1 + \frac 1n)^n \to  e &gt; 2 $$ 

$$ \exists\, n_0: \,\, \forall n &gt; n_0 \,\, a_{n+1} &gt; 2 a_n $$
$$  a_{n_0+p} &gt; 2^p a_{n_0}\rightarrow \lim_{n \to \infty} a_n = \infty \rightarrow \lim_{n \to \infty}\frac{n^{n+1}}{n!\,\, e^n} = \infty $$