calculus - Convergence of $sum_{n=1}^inftyfrac {n^{n}}{e^nn!}$

Check the convergence of: $\displaystyle\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$

Using the root test I get: $\displaystyle\lim_{n \to\infty} \dfrac {n}{e\sqrt[n]{n!}}$ now I'm left with showing that $n > \sqrt[n]{n!} \ \ \forall n$, can I just raise it to the power of $n$ like so: $\ n^n>n!$ ?

Alternatively, using limit arithmetic: $\displaystyle\lim_{n \to\infty} \dfrac {n}{e\sqrt[n]{n!}}=\displaystyle\lim_{n \to\infty} \dfrac {1}{\large\frac e n \sqrt[n]{\frac {n!}{n^n}}}>1$ (that's not very persuasive I know) so it diverges.

Edit: Root test won't work.

Note: Stirling, Taylor or integration are not allowed.