Find $\lim_{n\to \infty} n!\frac{e^n}{n^n}$ can anyone help me with this limit?
$$\lim_{n\to \infty} n!\frac{e^n}{n^n}$$
I know that the result is infinity but I can't find a way to prove it.
What I tried was to "split" the expression in half to have a pruduct of n/2 and n/2 fractions and then made it smaller by taking first n/2 fractions as $\frac{e}{n}\frac{n}{2}$ and the other n/2 fractions as $\frac{e}{n}$ leading to $\lim_{n\to \infty}(\frac{e}{n}\frac{n}{2})^\frac{n}{2}(\frac{e}{n})^\frac{n}{2}$ which equals zero so this way of using squeeze theorem doesn't work. L'Hospital rule didn't help me either.
Thank you for your advice or hints
 A: Suppose that $n$ is a positive integer. By the concavity of $\log$, we have
\begin{align*}
\log n! = & \sum\limits_{k = 1}^n {\log k}  \ge \sum\limits_{k = 1}^n {\int_{k - 1/2}^{k + 1/2} {\log xdx} }  = \int_{1/2}^{n + 1/2} {\log xdx} 
\\ &
 = \left( {n + \frac{1}{2}} \right)\log \left( {n + \frac{1}{2}} \right) - \left( {n + \frac{1}{2}} \right) + \frac{{1 + \log 2}}{2}
\\ &
 \ge \left( {n + \frac{1}{2}} \right)\log n - n .
\end{align*}
Thus,
$$
\log \left( {\frac{{n!e^n }}{{n^n }}} \right) \ge \frac{1}{2}\log n,
$$
i.e.,
$$
\frac{{n!e^n }}{{n^n }} \geq \sqrt n.
$$
This shows that the LHS tends to $+\infty$ as $n\to+\infty$.
A: Using Stirling's approximation we have that 
$$\sqrt{2\pi n} \leq \frac{n!e^n}{n^n}.$$
And since you lower bound diverges you have that limit diverges.
A: \begin{equation}\displaystyle\lim_{n\rightarrow\infty}\Bigg(\frac{n!e^n}{n^n}\Bigg) = \lim_{n\rightarrow\infty}\Bigg(\prod_{i=0}^{n-2}\Bigg(1-\frac{i+1}{n}\Bigg)\Bigg)e^n = \lim_{n\rightarrow\infty}\Bigg(\underbrace{1\times 1\times..\times1}_\text{n-1 times}\times\frac{1}{n}\times e^n\Bigg)=\lim_{n\rightarrow\infty}\Big(\frac{e^n}{n}\Big)=\lim_{n\rightarrow\infty}e^{n-1}=\infty\end{equation}
(by L'Hôpital's rule). To clarify,
\begin{equation}
\frac{n!}{n^n} = \frac{n(n-1)(n-2)...1}{\underbrace{n\times n\times n... \times n}_\text{n times}} = \frac{n}{n}\times\frac{n-1}{n}\times...\times\frac{1}{n} = 1\times\Bigg(1-\frac{1}{n}\Bigg)\times....\times\frac{1}{n}
\end{equation}
And,
$$\frac{1}{n} = \Bigg(1-\frac{n-1}{n}\Bigg)$$ which is the result when $i=n-2$. Thus, the limit diverges.
