I would like to show that $(1-e^{-x} )^2$ is approximated well near $x=0$ with $x^2$ via Taylor expansion but can't quite seem to complete the job.
I know that by expanding the exponential into its Taylor series and tidying some terms I end up with $(1-e^{-x})^2=(x-\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}...)^2$.
By raw calculation I can see that this ends up approximating $x^2$ nicely but I do not know how to show this explicitly.
$\begin{align*} (1-e^{-x})^2 &=1-2e^{-x}+e^{-2x}\\ &=1-2\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{n!}+\sum_{n=0}^{\infty} \frac{(-2)^nx^{n}}{n!}\\ &=1-\sum_{n=0}^{\infty} \frac{(2(-1)^n-(-2)^n)x^n}{n!}\\ &=1-\sum_{n=0}^{\infty} \frac{(-1)^n(2-2^n)x^n}{n!}\\ &=1-(1-0x-x^2+\sum_{n=3}^{\infty} \frac{(-1)^n(2-2^n)x^n}{n!})\\ &=x^2-\sum_{n=3}^{\infty} \frac{(-1)^n(2-2^n)x^n}{n!}\\ \end{align*} $