obscure limits

Proof

$\lim_{n\to\infty }\, n \sin ( 2\pi n! e)\,$	$=\lim_{n\to\infty}\, n \sin \left( 2\pi \sum_{k=0}^\infty \frac{n!}{k!} \right)\,$ ,
	$=\lim_{n\to\infty}\, n \sin \left( 2\pi \sum_{k={n+1}}^\infty \frac{n!}{k!} \right)\,$ ,
	$=\lim_{n\to\infty}\, n \sin \left( 2\pi \left [ \frac{1}{n+1} + \sum_{k={n+2}}^\infty \frac{n!}{k!} \right]\right)\,$ ,
	$=\lim_{n\to\infty}\, n \sin \left( 2\pi \left [ \frac{1}{n+1} + \sum_{k={n+2}}^\infty \frac{n!}{k!} \right]\right)\,$ ,
	$=\lim_{n\to\infty}\, n \sin \left( \frac{2\pi}{n+1} \right)\,$ ,
	$=2\pi\,$ .