Computing a limit involving $n^n$, $n!$, and an exponential

Question

I want to show that for arbitrary $\epsilon>0$, $$\lim_{n\to\infty}\frac{n^{n+1}}{n!} \text{exp}\left(-\left(n^\epsilon + n \right)\right)=0.$$ How can I go about proving this? A series expansion for the exponential doesn't seem to be helpful. I can't use Stirling's approximation for $n!$ because ultimately I am using this as one step in the proof of Stirling's approximation. Thank you.

Answer 1

Hint: Take the $\ln$ of the expression to get $$(n+1)\ln n - \ln n! - n^\epsilon -n.$$ Now use $\ln n! =\sum_{k=1}^n \ln k > \int_1^n \ln x\, dx,$ which should be clear after staring down some rectangles to give up their information. That integral can be evaluated exactly. Proceed ...