Is there a nice closed form for computing the n-th derivative of
$$f(x)=\frac{e^x}{x-1}$$ ?
I tried writing $xf(x)=e^x$ and then applying the Leibniz formula for the n-th derivative and this gives $$(x-1)f^{(n)}(x)+nf^{(n-1)}(x)=e^x$$ and here i arrive to a differential equation (which i am not very aware of the solving methods)
Another attempt: i observed by computing the first few derivatives that the form is $$f^{(n)}(x)=\frac{e^xp_{n}(x)}{(x-1)^{n+1}}$$ where $p_{n}$ is a polynomial of degree $n$ and satisfies $$p_{n+1}(x)=(x-n-2)p_{n}(x)+(x-1)p_{n}^{'}(x)$$ but again it seemes difficult to find $p_n$ explicitly if there really is a nice explicit formula.
Last try: i tied directly applying Leibniz formula for the product $$ e^x\frac{1}{x-1}$$ and i found that $$f^{(n)}(x)=e^x \{ C_{n}^{0}\frac{1}{x-1}-C_{n}^{1}\frac{1!}{(x-1)^2}+C_{n}^{2}\frac{2!}{(x-2)^3}-...+(-1)^{n}C_{n}^{n}\frac{n!}{(x-1)^{n+1}} \}$$ and here , i don't see if there is a nice closed form for all this sum.
All i see is that $C_{n}^{k}k!=A_{n}^{k}$ where i used the old notations $C_{n}^{k}$=binomial numbers (combinations) And $A_{n}^{k}$=number of ordered subsets with k elements for a set with n elements. Obs: By $f^{(n)}$ i denoted the n-th derivative of the function f.
So, am i missing something obvious? Or it is not that simple to find the closed form of the n-th derivative of this function?
