Answer
The series converges by the Ratio Test.
Work Step by Step
Use the Ratio Test: $\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$
Let $$a_n=\frac{n!}{{e^n}^2}$$
Then $$\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_\limits{n\to\infty}\left|\frac{(n+1)!}{{e^{(n+1)}}^2}\cdot\frac{{e^n}^2}{n!}\right| $$
We can take away the absolute value bars and then simplify.
$$ \lim_\limits{n\to\infty}\frac{(n+1)n!\cdot {e^n}^2}{e^{n^2+2n+1}n!}=\lim_\limits{n\to\infty}\frac{(n+1)}{e^{2n+1}}$$ so, as $n\to \infty$, $$\lim_\limits{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\to0 $$
thus, the series is convergent by the Ratio Test.