Compute $\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}$

By ratio test

$$\frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n}=\frac1{n+1}\left(1+\frac1n\right)^n\to 0$$

then

$$\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}=0$$

Another way. Note that as $n\to +\infty$, $$\frac{n^n}{(n!)^2}=\prod_{k=1}^n\frac{n}{ k(n+1-k)}\leq \frac{n}{ \lfloor\frac{n+1}{2}\rfloor (n+1-\lfloor\frac{n+1}{2}\rfloor)}\leq \frac{4}{n}\to 0$$ because each factor $\frac{n}{ k(n+1-k)}\leq 1$, and the smallest one is in the middle.