Push it to the limit

We'll use the well known identity $\lim_{n\rightarrow\infty} \left(1+\dfrac{k}{n}\right)^n=e^k$ Therefore $\lim_{n\rightarrow \infty} \prod_{i=1}^{2014}\left(1+\dfrac{i}{n}\right)^n=\prod_{i=1}^{2014}e^i=e^{\sum_{i=1}^{2014}i}=e^{2029105}$

The desired answer is, therefore, $\fbox{105}$ .

It is a well known fact that

$\displaystyle\lim_{n \to \infty} (1+\frac{i}{n})^n=e^i$

for some constant i.That can be proved by using the Taylor series for $e^x$ and expanding with the binomial formula and taking the limit as n approaches infinity.Then our product will be

$e.e^2.e^3....e^{2014}=e^{1007.2015}$ .Taking mod 1000 we get the answer $\boxed{105}$ .