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!

Is not full answer only for 1):
Note that:
$$\sum _{j=0}^{\infty } (-1)^j e^{j x}=\frac{1}{1+e^x}$$
so,
$$\color{red}{\frac{\partial ^k}{\partial x^k}\frac{1}{1+e^x}}=\sum _{j=0}^{\infty } \frac{\partial
   ^k\left((-1)^j e^{j x}\right)}{\partial x^k}=\sum _{j=0}^{\infty } (-1)^j e^{j x}
   j^k=\color{red}{\Phi \left(-e^x,-k,0\right)}$$
where $\Phi \left(-e^x,-k,0\right)$ is: Hurwitz-Lerch transcendent function.
