Divergent infinite series $n!e^n/n^n$ - simpler proof of divergence? [duplicate]

Note $$\frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=\frac{(n+1)e}{(n+1)^{n+1}/n^n}=\frac{e}{(n+1)^n/n}=\frac{e}{(1+1/n)^n}>1$$ since $e>(1+1/n)^n$ for all $n$. It follows that $$\frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>\frac{n!e^n}{n^n},$$ so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.

Now, perhaps your teacher meant $\frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $n\cdot\frac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $\frac{n^n}{n!e^n}\geq\frac{1}{en}$, and by divergence of harmonic series this series diverges too.