Taylor approximation $f(x)=e^{-e^x}$ for $\int_0^1 f(x)dx$ with an error less than $10^{-2}$ Taylor approximation of $f(x)=e^{-e^x}$ for $\displaystyle\int_0^1 f(x)dx$ with an error less than $10^{-2}$ from $g(x)=e^x$ using substitution.
This is what I did for getting the Taylor series of $e^{-e^x}$:
$ e^{\displaystyle-e^x}=e^{-\displaystyle\sum_{k=0}^n\frac{x^k}{k!}}=\displaystyle \prod_{k=0}^n\sum_{j=0}^n\frac{(-1)^j}{j!}\bigg(\frac{x^k}{k!}\bigg)^j$.
I'm not sure how to compute the error, using Lagrange Reminder, or something that helps. I'm stuck because of the productory.
 A: I totally understand your difficulties with such an approach.
Let $y=e^x$ and develop $e^{-y}$ as an infinte series around $y=1$ (corresponding to $x=0$). Replace $y$ by $e^x$ to have
$$f(x)=e^{-e^x}=\sum_{n=0}^\infty (-1)^n\frac{ e^{n x}}{n!}$$ and
$$\int_0^1 f(x)\,dx=\sum_{n=0}^\infty (-1)^n\frac{ 1}{n!}\int_0^1 e^{n x}\,dx=\sum_{n=0}^\infty (-1)^n\frac{ 1}{n!}\frac{e^n-1}{n}$$ So, if you make the summation up to $p$, what you want is that
$$R_p=\frac{e^{p+1}-1}{(p+1)^2 \,p!} \leq 10^{-k}$$ Neglecting the $-1$ in numerator, we then want
$$(p+1)^2\,p! \geq e^{p+1}\, 10^k$$ We shall simplify noting that $(p+1)^2\,p!\sim (p+2)!$ and write
$$(p+2)! \geq e^{p+2}\, \frac{10^k} e$$
Looking at this question of mine and adapting @robjohn's magnificent approximation to this problem, we have
$$p\sim e^{2+W(t)}-\frac 52 \qquad \text{where} \qquad t=\frac{2 k\log \left(10\right)-3-\log (2 \pi )}{2 e^2}$$ $W(t)$ being Lambert function.
Trying for $k=2$  gives $p=6.83823$ than $\lceil p \rceil=7$. Notice that, without any approximation, the exact result would be $p=6.93724$.
Checking with the exact $R_p$
$$R_{6}=0.0311 > 10^{-2}\qquad \text{but} \qquad  R_{7}=0.0092<10^{-2}$$
