TITLE: Elementary proof for $\lim\limits_{n \to\infty}\frac{n!e^n}{n^n} = +\infty$
QUESTION [12 upvotes]: 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.)

REPLY [3 votes]: 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.