Higher-order derivative of $1/(1+e^x)$ Let 
$$
f(x) = \frac{1}{1+e^x}
$$
1) Is there a closed formula for the $k$-derivative $f^{(k)}(x)$?
2) Is there a simple argument which shows for $k \geq 1$ that
$$
\lim_{x \to \pm \infty} f^{(k)}(x) = 0 ?
$$
Thanks!

Write $e^x=:u$ for short. Then there is a sequence $(p_n)_{n\geq0}$ of polynomials such that
$$f^{(n)}(x)={p_n(u)\over(1+u)^{n+1}}\qquad(n\geq0)\ .$$
It is easy to show that the $p_n$ satisfy the recursion
$$p_0=1,\qquad p_{n+1}=(u+u^2)\, p_n'-(n+1) u\, p_n\qquad(n\geq0)\ .$$
E.g., one obtains
$$p_6(u)=u (-1+57 u-302 u^2+302 u^3-57 u^4+u^5)\ .$$
