Simplified form for $\frac{\operatorname d^n}{\operatorname dx^n}\left(\frac{x}{e^x-1}\right)$? I have found the following formula:
$$\frac{\operatorname  d^n}{\operatorname dx^n}\left(\frac{x}{e^x-1}\right)=(-1)^n\,\frac{n\sum\limits_{k=0}^{n}e^{kx}\sum\limits_{i=0}^{k}(-1)^i\binom{n+1}{i}(k-i)^{n-1}+x\sum\limits_{k=0}^{n}e^{kx}\sum\limits_{i=0}^{k}(-1)^i\binom{n+1}{i}(k-i)^n}{\left(e^x-1\right)^{n+1}}.
$$
My proof of this formula is complicated.
Can somebody find some simple proof?
 A: Here's a starting point.
Using the general Leibniz rule, 
$$\frac{d^n}{dx^n}\frac{x}{e^x-1}=\sum_{k=0}^n\binom{n}{k}\left(\frac{d^k}{dx^k}x\right)\left(\frac{d^{n-k}}{dx^{n-k}}\frac{1}{e^x-1}\right).$$
Now, the "0-th" derivative of $x$ is $x$, and the 1st derivative of $x$ is $1$, and all 2nd and higher order derivatives vanish. Then, the only non-zero terms in the sum above are the k=0,1 terms, so we get:
$$\frac{d^n}{dx^n}\frac{x}{e^x-1} = \binom{n}{0}\left(\frac{d^0}{dx^0}x\right)\left(\frac{d^{n-0}}{dx^{n-0}}\frac{1}{e^x-1}\right)+\binom{n}{1}\left(\frac{d^1}{dx^1}x\right)\left(\frac{d^{n-1}}{dx^{n-1}}\frac{1}{e^x-1}\right)\\
 = x\left(\frac{d^{n}}{dx^{n}}\frac{1}{e^x-1}\right) + n\left(\frac{d^{n-1}}{dx^{n-1}}\frac{1}{e^x-1}\right).$$
At the very least this should obviate the need for complicated products of summations as you currently have.
A: i hope this would help you in some way
$$\frac{x}{e^x-1}=\frac{x}{\cos(ix)-i\sin (ix)-1}$$
$$=\frac{x}{-(1-\cos (ix))-\sin (ix)}$$
$$=\frac{x}{-2\sin^2(\frac{ix}{2})-\sin (ix)}$$
$$=\frac{x}{-2\sin (\frac{ix}{2})(\sin (\frac{ix}{2})+\cos (\frac{ix}{2}))}$$
$$=\frac{x}{-2\sin (\frac{ix}{2})e^{\frac{-x}{2}}}$$
$$=\frac{xe^{\frac{x}{2}}}{i\sinh (\frac{x}{2})}$$
A: Related problems: (I), (II). Here is a formula 

$$ \sum _{k=0}^{n}  \sum _{m=0}^{n-k} ( -1 )^{m} m! {n\choose k} {n-k \brace m}(2-k)_k\,\frac{{x}^{1-k}\,{\rm e}^{mx}}{ \left( {{\rm e}^{x}}-1 \right)^{m+1}}, $$

where $(a)_b$ is the Pocchammer symbol and $ {n \brace k} $ is the Stirling numbers of the second kind.
A: The Taylor series of $\frac{x}{e^x-1}$ is well-known:
$$\frac{x}{e^x-1}=\sum_{k=0}^\infty \frac{B_k}{k!} x^k$$
$$\Longrightarrow \frac{\operatorname d^n}{\operatorname dx^n}\left(\frac{x}{e^x-1}\right)=\sum_{k=0}^\infty \frac{B_k}{k!} \frac{\operatorname d^n}{\operatorname dx^n}x^k$$
$$=\sum_{k=0}^\infty \frac{B_k}{k!} \frac{k!}{(k-n)!}x^{k-n}=\sum_{k=0}^\infty  \frac{B_k}{(k-n)!}x^{k-n}.$$
