Elementary proof for $\lim\limits_{n \to\infty}\frac{n!e^n}{n^n} = +\infty$ As seen in this question, the series $\sum\limits_{n=1}^{\infty} \dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)
However, more is true: not only are the terms greater than $1$, they blow up to $+\infty$! To see this, one can apply Stirling's approximation and get $\dfrac{n!e^n}{n^n} \sim \sqrt{2\pi n}$.
My question is: is there an elementary proof for $\lim_{n \to\infty}\dfrac{n!e^n}{n^n} = +\infty$?
The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+\infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.
(To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
 A: One can use integrals. The work is a little long, but quite rewarding. Consider $$I_n=\int_0^{\pi /2}\sin^{n}tdt$$
Then $$I_{2n+1}\leqslant I_{2n}\leqslant I_{2n-1}$$
On the other hand integrating by parts gives 
$${I_{2n + 1}} = \frac{{2n}}{{2n + 1}}{I_{2n - 1}}$$
It follows that $$1\leqslant \frac{{{I_{2n}}}}{{{I_{2n + 1}}}}\leqslant \frac{{{I_{2n - 1}}}}{{{I_{2n + 1}}}}$$ so $$\frac{{{I_{2n}}}}{{{I_{2n + 1}}}}\to 1$$
We proceed to evaluate that limit in a different way. Integrating by parts like before gives us $${I_{2n}} = \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}\frac{\pi }{2}$$ $${I_{2n + 1}} = \frac{{\left( {2n } \right)!!}}{{\left( {2n+ 1} \right)!!}}$$
Thus, we find $$\frac{{{I_{2n + 1}}}}{{{I_{2n}}}} = \frac{{\left( {2n} \right)!{!^2}}}{{\left( {2n - 1} \right)!!\left( {2n + 1} \right)!!}}\frac{2}{\pi } \to 1$$ or $$\prod\limits_{k = 1}^\infty  {\frac{{4{k^2}}}{{4{k^2} - 1}}}  = \frac{\pi }{2}$$
This is the celebrated Wallis product for $\pi$. Now onto Stirling. Consider the limit $$A = \mathop {\lim }\limits_{n \to \infty } \frac{{n!{e^n}}}{{{n^{n + 1/2}}}}$$ Then one must have by $n\mapsto 2n$ that  $$1 = \frac{A}{A} = \mathop {\lim }\limits_{n \to \infty } \frac{{n!{e^n}}}{{{n^{n + 1/2}}}}\frac{{{{\left( {2n} \right)}^{2n + 1/2}}}}{{\left( {2n} \right)!{e^{2n}}}} = \sqrt 2 \mathop {\lim }\limits_{n \to \infty } \frac{{{n^n}}}{{{e^n}n!}}\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}$$
We now use Wallis product formula. Squaring, this means that $$1 = 2{\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} \right)^2}\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{2n + 1}}{n}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!!\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!\left( {2n + 1} \right)!!}}} \right)$$  And we thus get  Stirling's $$\frac{1}{{\sqrt {2\pi } }} = { {\mathop {\lim }\limits_{n \to \infty } \frac{{{n^{n + 1/2}}}}{{{e^n}n!}}} }$$ 
ADD  Denote by $a_n$ our sequence above. Then $${a_n} > {a_{n + 1}}$$ is equivalent to $${\left( {1 + \frac{1}{n}} \right)^{n + 1/2}} > e$$ under rearranging.  It remains to prove this.
A: Similar to robjohn's explanation, one can approach this using the so-called "trapezoidal rule" for estimating integrals. 
Thus, consider the integral $\int_1^n \log(x) \; dx$. The trapezoidal rule estimate gives, over each interval $[m, m+1]$, the estimate 
$$\int_m^{m+1} \log(x) \; dx \;\; = \;\; \frac1{2}(\log(m) + \log(m+1)) + e_m$$ 
where the error term $e_m$ is bounded above by a constant times $m^{-2}$. Summing from $m=1$ to $n-1$, we arrive at 
$$x\log(x) - x\; |_1^n = \int_1^n \log(x)\; dx \;\; = \;\; \frac1{2}\log(n)+ \sum_{1}^{n-1} \log(n) + E_n$$ 
where the error term $E_n = \sum_{m=1}^{n-1} e_m$ is some constant $E = \sum_{m=1}^\infty e_m$ minus an amount that is bounded above by some constant $C$ times $n^{-1}$. We may exponentiate to get $n^n e^{-n}e^{-1} = e^{E_n}(\frac{n!}{\sqrt{n}})$, which may be massaged into the asymptotic formula 
$$\frac{e^n n!}{n^n} = (1 + O(n^{-1}))e^{1-E}\sqrt{n}$$ 
where $E$ could be determined if we want, but this gets the job done. See Terry Tao's post here. 
A: Considering the ratio of consecutive elements of
$$
a_n=\frac{n!e^n}{n^n}\tag{1}
$$
we get
$$
\begin{align}
\frac{a_n}{a_{n-1}}
&=\frac{(n-1)^{n-1}e}{n^{n-1}}\\
&=e\left(1-\frac1n\right)^{n-1}\tag{2}
\end{align}
$$
Taking the log of this ratio gives
$$
\begin{align}
\log\left(\frac{a_n}{a_{n-1}}\right)
&=1+(n-1)\log\left(1-\frac1n\right)\\
&=1-(n-1)\left(\frac1n+\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)\\
&=\frac1{2n}+O\left(\frac1{n^2}\right)\tag{3}
\end{align}
$$
Summing $(3)$ yields
$$
\log(a_n)=\frac12H_n+C+O\left(\frac1n\right)\tag{4}
$$
Since the harmonic series diverges, $(4)$ shows that $\log(a_n)$, and hence $a_n$, grows without bound.
