Evaluate $\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})$ 
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. 
 A: $$\sin (\pi \sqrt{(n!)^2 -n!} )=\sin (\pi \sqrt{(n!)^2 -n!} -\pi\cdot n! )=\sin \left(\pi \frac{-n!}{\sqrt{(n!)^2 -n!} +n!}\right)=\sin \left(\pi \frac{-1}{\sqrt{1 -\frac{1}{n!}} +1}\right)\to -\sin\frac{\pi}{2} =-1$$
A: There are three different answers with 2 different solutions at the end, so my take at it would be the following:
The sequence is a subsequence of $2u$ for $u\in\mathbb{N}$ (as factorial is even), thus we get
$$\lim_{n\to\infty}\sin^2\left(\pi\sqrt{n!^{2}-n!}\right)=\lim_{2u\to\infty}\sin^2\left(\pi \sqrt{(2u)^2-2u}\right)=\lim_{2u\to\infty}\sin^2\left(\pi 2u\sqrt{1-\frac{1}{2u}}\right)$$
And expanding $\sqrt{x}$ to a power series around $x=1$
  gives us:
$$T\left(x\right)=1+f'\left(1\right)\left(x-1\right)+o\left(x-1\right)^{2}=1+\frac{1}{2}\left(x-1\right)+o\left(x-1\right)^{2}$$
Hence at $x=1-\frac{1}{2u}$ we get 
$$T\left(\frac{1}{2u}\right)=1-\frac{1}{2}\cdot\frac{1}{2u}+o\left(\frac{1}{2u}\right)^{2}$$
And $$\sin^2\left(\pi 2u\sqrt{1-\frac{1}{2u}}\right)=\sin^2\left(\pi\cdot 2u\left(1-\frac{1}{2}\cdot\frac{1}{2u}+o\left(\frac{1}{2u}\right)^{2}\right)\right)=\sin^2\left(-\frac{\pi}{2}+o\left(\frac{1}{2u}\right)^2\right)$$
Thus taking $2u$ to $\infty$ gives the limit of $(-1)^2=1$.
A: In fact
\begin{eqnarray}
\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})&=&\lim \limits_{n\to \infty }\sin^2 \pi[n!- \sqrt{(n!)^2-(n!)}]\\
&=&\lim \limits_{n\to \infty }\sin^2 \pi\frac{n!}{n!+ \sqrt{(n!)^2-(n!)}}\\
&=&\lim \limits_{x\to \infty }\sin^2 \pi\frac{x}{x+ \sqrt{x^2-x}}\\
&=&\sin^2\frac{\pi}{2}\\
&=&1.
\end{eqnarray}
