Weighted uniform convergence of Taylor series of exponential function 
Is the limit 
  $$
e^{-x}\sum_{n=0}^N \frac{(-1)^n}{n!}x^n\to e^{-2x}  \quad \text{as } \ N\to\infty \tag1
$$
  uniform on $[0,+\infty)$? 

Numerically this appears to be true: see the difference of two sides in (1)  for $N=10$ and $N=100$ plotted  below. But the convergence is very slow (logarithmic  error $\approx N^{-1/2}$ as shown by Antonio Vargas in his answer). In particular, putting $e^{-0.9x}$ and $e^{-1.9x}$ in (1) clearly makes convergence non-uniform. 
One difficulty here is that the Taylor remainder formula is effective only up to $x\approx N/e$, and the maximum of the difference is at $x\approx N$.


The question is inspired by an attempt to find an alternative proof of $\epsilon>0$ there is a polynomial $p$ such that $|f(x)-e^{-x}p|<\epsilon\forall x\in[0,\infty)$. 

I'd like to provide another solution which is a mixture of Antonio's and Landscape's. One can also write $$\left|r_n(x)\right|=\int_0^x {e^{-t}}\frac{(x-t)^n}{n!}dt$$
by virtue of Taylor's theorem with the integral remainder. But then again $$\left| {{r_n}(x)} \right| \leqslant \int_0^x {\frac{{{{(x - t)}^n}}}{{n!}}dt}  = \int_0^x {\frac{{{u^n}}}{{n!}}du}  = \frac{x^{n + 1}}{{\left( {n + 1} \right)!}}$$
and Strling does the job.
