This seems like a weird question, because the series has good convergence and there's no need to use other methods to estimate it for $N \to \infty$.
However, after seeing this question, I tried to come up with some approximation which could allow us to treat integrals containing this sum in the denominator.
Regardless of the particular application, I just wanted to ask if there's a way to approximate this sum using a finite combination of elementary or special functions?
Not counting the obvious $e^x-\sum_{k=0}^{N-1} \frac{x^k}{k!}$ of course.
Euler-Maclaurin formula doesn't seem very promising, because the resulting integral is even more complicated $$\int_N^\infty \frac{x^y}{\Gamma(y+1)}dy$$ And the derivatives containing various combinations of polygamma functions are also a pain to write.
If there's no better asymptotic expression than the sum itself, so be it.
Best Answer
Note that $$ \frac{n!}{x^n}\sum_{k=n}^\infty\frac{x^k}{k!} =1+\sum_{k=n+1}^\infty\frac{x^{k-n}n!}{k!}\\ $$ and for $n\ge|x|$, $$ \begin{align} \left|\sum_{k=n+1}^\infty\frac{x^{k-n}n!}{k!}\right| &\le\sum_{k=1}^\infty\frac{|x|^k}{(n+1)^k}\\ &=\frac{|x|}{n+1-|x|} \end{align} $$ Thus, $$ \sum_{k=n}^\infty\frac{x^k}{k!} =\frac{x^n}{n!}\left(1+O\!\left(\frac1n\right)\right) $$