Is there a relatively small $c$ so that $e^x-\sum\limits_{n=0}^{c \cdot x} \frac{x^n}{n!} < 1$? What is the precise number of terms needed in Taylor expansion of $e^x$ in order to achieve a precision of $1$?
It seems that we simply need something like $e \cdot x$ terms or about that, but I am not completely sure that this estimation comes from Taylor series error.
I try
$$R_n(x)=\frac{e^cx^n}{n!}<1$$
taking the worst case scenario $c=x$
$$e^x x^n < n!$$
$$x+n\ln(x) < n\ln(n)-n$$
$$x< n\ln(\frac{n}{x})-n$$
$$n=kx$$
$$x< kx\ln(\frac{kx}{x})-kx$$
$$k(\ln(k)-1) > 1$$
$$k > e^{W(\frac{1}{e})+1} \approx 3.59112$$
Taking into account how precise $\ln(n!) \approx n\ln(n)-n$ is, we indeed should have something like linear dependency.
Is this the best possible estimation we can have? Is it even correct to ask first actually?
 A: A bound on the remainder terms by a geometric sum gives that for $N>2x$ the error of the partial sum is smaller than twice the next term. This would give the condition
$$
\frac{x^N}{N!}<\frac12
$$
for which it is sufficient to have by the Stirling formula (and as $\sqrt{2\pi}>2$)
$$
x^N<\left(\frac Ne\right)^N\iff x<\frac Ne \iff N>ex.
$$
As $e>2$, this satisfies the assumptions of the bound $N>2x$ used in the error estimate.
Again by the full Stirling formula, the error estimate can be refined to
$$\approx2\frac{x^N}{\sqrt{2\pi N}(\frac{N}e)^N}<\frac1{\sqrt{N}}\left(\frac{ex}N\right)^N<\frac1{\sqrt{N}}.$$

The remainder formula is for $0<x<N/2$
$$
\sum_{n=N}^\infty\frac{x^n}{n!}\le\frac{x^N}{N!}\sum_{k=0}^\infty\frac{x^k}{(N+1)^k}=\frac{x^N}{N!}\frac1{1-\frac{x}{N+1}}<2\frac{x^N}{N!}
$$
A: Hint
Given
$$
e^{\,x}  - \sum\limits_{n = 0}^{c\,x} {{{x^{\,n} } \over {n!}}}  < 1\quad  \Leftrightarrow \quad 1 - e^{\, - x}  < e^{\, - x} \sum\limits_{n = 0}^{c\,x} {{{x^{\,n} } \over {n!}}} 
$$
it is known that the partial sum of the $e^x$  power series  divided by $e^x$ is given by
the Upper Incomplete Regularized Gamma Function.   
So, from
$$
\eqalign{
  & e^{\,x}  - \sum\limits_{n = 0}^{c\,x} {{{x^{\,n} } \over {n!}}}  < 1\quad  \Leftrightarrow \quad
 1 - e^{\, - x} \sum\limits_{n = 0}^{c\,x} {{{x^{\,n} } \over {n!}}}  < e^{\, - x} \quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad 1 - Q(cx + 1,x) = P(cx + 1,x) < e^{\, - x}  \cr} 
$$
you might find other approaches to solve your problem
A: Thanks to all.
The answer is maybe tricky but indeed possible.
All we need is to estimate the implicit function $y(x)$ in form of
$$ e^x(1-\frac{\Gamma(1+xy(x),x)}{\Gamma(1+xy(x))}) = 1 $$
$$\frac{\ln(1-\frac{\Gamma(1+xy(x),x)}{\Gamma(1+xy(x))})}{x} = -1 $$
This is indeed giving something like $y(x) \to e$ or around that number at infinity.
So we need to prove something down the line, if it is really $e$:
$$ \lim_{x \to \infty} \frac{\ln(1-\frac{\Gamma(1+ex,x)}{\Gamma(1+ex)})}{x} = -1 $$
and we are done.
Maybe using
$$ \Gamma(s,x) = \Gamma(s) - \gamma(s,x) $$
and
$$ s\gamma(s,x) \sim x^s e^{-x} $$
might help.
$$ \lim_{x \to \infty} \frac{\ln(\frac{e^{-x}x^{ex+1}}{(ex+1)\Gamma(1+ex)})}{x} = -1 $$
which requires
$$ \lim_{x \to \infty} \frac{\ln(\frac{x^{ex+1}}{(ex+1)\Gamma(1+ex)})}{x} = 0 $$
and this is correct taking the asymptotic behavior of $\Gamma(x+1) \sim \sqrt{2 \pi x}\left(\frac{x}{e}\right)^x$ into account because $x^{ex}$ is canceled.
Notice that $c$ must be greater or equal than $e$ otherwise $\left(\frac{z}{e}\right)^z$ in the asymptotic expression for $\Gamma(1+z)$ would make the limit being different from $0$ for $z=cx, c \neq e$
Job done!
A very sensitive limit though
$$ \lim_{x \to \infty} \frac{\ln(\frac{x^{x+1}}{(x+1)\Gamma(1+x)})}{x} = 1 $$
$$ \lim_{x \to \infty} \frac{\ln(\frac{x^{2x+1}}{(2x+1)\Gamma(1+2x)})}{x} = 2-\ln(4) $$
Using the shortest possible expression and regularized gamma function $\operatorname{P}(a,z)$ we have this implicit definition of our function
$$\operatorname{P}(xy(x)+1,x)=e^{-x}$$
A: Starting from @LutzL's answer, we need compute $n$ such that
$$n!=2x^n$$
If you look at this question of mine, @Robjohn provided a splendid approximation.
Making in the answer $k=\frac{\log (2)}{\log (10)}$ and $a=x$, this would make
$$n\sim e x \exp\Bigl[W\left(-\frac{1}{2 e x}\log \left(\frac{\pi  x}{2}\right)\right) \Bigl]-\frac 12$$ which would make for large enough value of $x$
$$c \sim   \exp\Bigl[1+W\left(-\frac{1}{2 e x}\log \left(\frac{\pi  x}{2}\right)\right) \Bigl]$$ which, for sure, tends to $e$ (from below).
