general formula for the nth derivative of $f(x)=\frac{1}{1+e^{x}}$ consider the function :$$f(x)=\frac{1}{1+e^{x}}$$
the nth derivative of the function is given by the following formula:
$$f^{(n)} (x)=\sum_{k=1}^{n+1}a_{n,k}\frac{1}{\left(1+e^{x}\right)^{k}}$$ where
$$a_{n,k}=\left(-1\right)^{n}\sum_{j=0}^{k-1}\left(-1\right)^{j}{{k-1}\choose{j}}\left(j+1\right)^{n}$$
my question is that:how the formula can be derived without using induction? I have no idea about that, so any hint or full proof would be highly appreciated.
 A: A calculation without recurrence relation and without induction.
Let’s use the following formula, both sides come from the mixed bivariate generating function $~\displaystyle e^{z(e^x-1)}~$ for the Stirling numbers of the second kind:

$$e^{-z}\sum\limits_{k=0}^\infty\frac{z^k}{k!}k^n = \sum\limits_{k=0}^n z^k S(n,k)$$

$S(n,k)~$ are the Stirling numbers of the second kind .
Setting $~z:=at~$, multiplicating by $~e^{-t}~$ and integrating from $~t=0~$ to $~\infty~$ we get: 

$$\sum\limits_{k=0}^\infty k^n\left(\frac{a}{1+a}\right)^k = (a+1)\sum\limits_{k=0}^n a^k k!S(n,k)$$

Setting $~\displaystyle\frac{a}{1+a} = -e^{-x}~$, substracting $~S_{n,0}=0^n~$ and 
using $~S(n+1,k+1)=(k+1)S(n,k+1)+S(n,k)~$ we get:

$$\sum\limits_{k=1}^\infty k^n\left(-e^{-x}\right)^k = \sum\limits_{k=0}^n \frac{(-1)^{k+1} k!S(n+1,k+1)}{(1+e^x)^{k+1}}$$

It follows:
$\hspace{1cm}~\displaystyle\frac{d^n}{dx^n}\frac{1}{1+e^x} = (-1)^{n+1}\sum\limits_{k=1}^\infty k^n\left(-e^{-x}\right)^k = \sum\limits_{k=0}^n \frac{a_{n,k+1}}{(1+e^x)^{k+1}}~$ 
with $~~~\displaystyle a_{n,k+1} = (-1)^{n+k} k!S(n+1,k+1) = \left(-1\right)^{n}\sum_{j=0}^{k}\left(-1\right)^{j}{{k}\choose{j}}\left(j+1\right)^{n}$

Notes:
$\displaystyle e^{z(e^x-1)}=e^{-z}e^{ze^x}=e^{-z}\sum\limits_{k=0}^\infty\frac{z^k}{k!}e^{xk}=e^{-z}\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{n=0}^\infty\frac{x^n k^n}{n!}=\sum\limits_{n=0}^\infty\frac{x^n }{n!}\left(e^{-z}\sum\limits_{k=0}^\infty\frac{z^k k^n}{k!}\right)$
$\displaystyle e^{z(e^x-1)}=\sum\limits_{k=0}^\infty\frac{z^k}{k!}(e^x-1)^k=\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{j=0}^k(-1)^{k-j}{\binom k j}e^{xj}=$
$\displaystyle\hspace{1.5cm}=\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{j=0}^k(-1)^{k-j}{\binom k j}\sum\limits_{n=0}^\infty\frac{x^n}{n!}j^n=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\left(\sum\limits_{k=0}^\infty\frac{z^k}{k!}\sum\limits_{j=0}^k(-1)^{k-j}{\binom k j}j^n\right)$
$\displaystyle\hspace{1.5cm}=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\left(\sum\limits_{k=0}^\infty z^k S(n,k)\right)=\sum\limits_{n=0}^\infty\frac{x^n}{n!}\left(\sum\limits_{k=0}^n z^k S(n,k)\right)$ 
$\hspace{1.8cm}$ because of $~S(n,k)=0~$ for $~k>n$
Comparing the coefficients of $~x^n~$ we get the first formula.
And the second formula comes from integrating by $~\int\limits_0^\infty ... dt~$ ; 
with $~c>0~$ it's $\displaystyle~\int\limits_0^\infty\frac{t^n}{e^{ct}}dt=\frac{1}{c^{n+1}}\int\limits_0^\infty\frac{(ct)^n}{e^{ct}}d(ct)=\frac{n!}{c^{n+1}}~$ :
$\displaystyle\sum\limits_{k=0}^\infty\frac{(at)^k}{e^{at}k!}k^n = \sum\limits_{k=0}^n (at)^k S(n,k) ~~~~ |\cdot e^{-t} ~~~~ |\int\limits_0^\infty ... dt$
$\displaystyle\sum\limits_{k=0}^\infty\frac{a^k k^n}{k!}\int\limits_0^\infty\frac{t^k}{e^{(a+1)t}}dt = \sum\limits_{k=0}^n a^k S(n,k)\int\limits_0^\infty\frac{t^k}{e^t}dt$
$\displaystyle\sum\limits_{k=0}^\infty \frac{a^k k^n}{(1+a)^{k+1}} = \sum\limits_{k=0}^n a^k k!S(n,k) ~~~~ |\cdot (1+a)$
From there to the third formula: 
$\displaystyle\sum\limits_{k=0}^\infty k^n\left(\frac{a}{1+a}\right)^k = (a+1)\sum\limits_{k=0}^n a^k k!S(n,k)$
$\displaystyle = \sum\limits_{k=0}^n a^{k+1} k!S(n,k) + \sum\limits_{k=0}^n a^k k!S(n,k) = \sum\limits_{k=1}^{n+1} a^k (k -1)!S(n,k-1) + \sum\limits_{k=0}^n a^k k!S(n,k) $
$\displaystyle = a^{n+1} n!S(n,n) + \sum\limits_{k=1}^{n} a^k (k -1)!(S(n,k-1) + k S(n,k)) + a^0 0!S(n,0) $
$\displaystyle = a^{n+1} n!S(n,n) + \sum\limits_{k=1}^{n} a^k (k -1)!S(n+1,k) + S(n,0) $
$\displaystyle = \sum\limits_{k=1}^{n+1} a^k (k -1)!S(n+1,k) + S(n,0) = \sum\limits_{k=0}^{n} a^{k+1} k!S(n+1,k+1) + S(n,0) $
Substracting $~0^n=S(n,0)~$ leads to $\displaystyle \sum\limits_{k=1}^\infty k^n\left(\frac{a}{1+a}\right)^k = \sum\limits_{k=0}^{n} a^{k+1} k!S(n+1,k+1)~$ .  
With setting $\displaystyle ~\frac{a}{1+a}=-e^{-x}~$ we get $\displaystyle ~a=-\frac{1}{1+e^x}~$ and therefore the last formula. 
