It isn't hard to prove that:
$\int_0^x e^{-t} {t^n} dt = n! \cdot e^{-x}\left( e^x-\sum_{k=0}^{n} \frac{x^k}{k!}\right)$
Or put in a different way:
$\int_0^x e^{-t} \frac{t^n}{n!} dt = e^{-x}\left( e^x-\sum_{k=0}^{n} \frac{x^k}{k!}\right)$
Now, if we take the limit as $n\to \infty$ on both sides we will get:
$\mathop {\lim }\limits_{n \to \infty } \int_0^x e^{-t} \frac{t^n}{n!} dt = \mathop {\lim }\limits_{n \to \infty }e^{-x}\left( e^x-\sum_{k=0}^{n} \frac{x^k}{k!}\right)$
But $\mathop {\lim }\limits_{n \to \infty }\sum_{k=0}^{n} \frac{x^k}{k!}= e^x$
Thus the RHS side is necessarily zero.
This implies that
$\mathop {\lim }\limits_{n \to \infty } \int\limits_0^x {{e^{ - t}}\frac{{{t^n}}}{{n!}}dt} = 0$
The idea of the nullity of the limit is that as $n$ increases the "bump" of the function $y=\displaystyle e^{-x} \frac{x^n}{n!}$ tends to go further away of the origin, thus for any finite value of $x$ we can choose $N \geq n$ such that the "bump" of the function is sufficiently far from the interval $(0,x)$ that is it made insignificant.
However, I'd like to know if there is a proof for this.