Does this limit exist finitely? Find the value of the following (if it exists):

$$\lim_{n\to\infty}n\sin(2\pi en!)$$

Does it exist? I think that it doesn't exist but I can't prove it. Please help me.

Here is a complete proof:
For given $n\geq2$ one has
$$e\cdot n!=n!\sum_{t=0}^\infty{1\over t!}=n!\left(\sum_{t=0}^n{1\over t!}+\sum_{t=n+1}^\infty{1\over t!}\right)=m_n+r_n$$
with $m_n\in{\mathbb Z}$ and
$${1\over n+1}<r_n={1\over n+1}+{1\over (n+1)(n+2)}+\ldots<{1\over n}+{1\over n^2}+\ldots<{1\over n-1}\ .$$
Since 
$$a_n:=n\>\sin\left(2\pi\cdot e\cdot n!\right)=n\>\sin(2\pi r_n)=n\ \ 2\pi r_n\ {\sin(2\pi r_n)\over 2\pi r_n}$$
and $r_n\to 0$ it follows that
$$\lim_{n\to\infty}a_n=2\pi\lim_{n\to\infty}\bigl(n\> r_n\bigr)=2\pi\ .$$
