algebra precalculus - Evaluate $lim limits_{nto infty }sin^2 (pi sqrt{(n!)^2-(n!)})$

Evaluate $$\lim \limits_{n\to \infty }\sin^2 \left(\pi \sqrt{(n!)^2-(n!)}\right)$$

I tried it by Stirling's Approximation
$$n! \approx \sqrt{2\pi n}.n^n e^{-n}$$
but it leads us to nowhere.

Any hint will be of great help.

Answer

$$\sin (\pi \sqrt{(n!)^2 -n!} )=\sin (\pi \sqrt{(n!)^2 -n!} -\pi\cdot n! )=\sin \left(\pi \frac{-n!}{\sqrt{(n!)^2 -n!} +n!}\right)=\sin \left(\pi \frac{-1}{\sqrt{1 -\frac{1}{n!}} +1}\right)\to -\sin\frac{\pi}{2} =-1$$