Coefficient of $x^k$ in ${(1 - e^x)}^{-n}$

Question

What is the coefficient of $x^k$ in ${(1 - e^x)}^{-n}$?

This is what I tried-

Using negative binomial expansion and Taylor series expansion of $e^t$, $${(1 - e^x)}^{-n} = \sum_{i=0}^{\infty} {n+i-1 \choose i} e^{ix} = \sum_{i=0}^{\infty} {n+i-1 \choose i} \sum_{t=0}^{\infty} \frac{i^t x^t}{t!} = \sum_{t=0}^{\infty} \frac{x^t}{t!} \sum_{i=0}^{\infty} i^t {n+i-1 \choose i} $$ But I couldn't proceed further. Does a closed form for the coefficient of $x^k$ exists?

Also, I would like to calculate the value of the first $k$ coefficients of the above expansion. Is there an efficient algorithm for it?

Answer 1

Note that $$\frac{x}{e^x-1}=\sum_{j=0}^{\infty} B_j \frac{x^j}{j!}$$ where $$B_j=\sum_{m=0}^j\frac{1}{m+1}\sum_{i=0}^{m}(-1)^i\binom mi i^j$$ is the $j$-th Bernoulli number. Hence $$[x^k]{(1 - e^x)}^{-n}=(-1)^n[x^{k+n}]\left(\frac{x}{e^x-1}\right)^n =(-1)^n\sum_{j_1+\dots+j_n=n+k}\frac{B_{j_1}}{j_1!}\cdots \frac{B_{j_n}}{j_n!}$$ where $j_1,\dots,j_n$ are non-negative integers.

P.S. For example for $n=3$ and for $k=0,1,2,3,4,5$, the coefficients are: $$\frac{3}{8},-\frac{19}{240},\frac{1}{160},\frac{1}{945},-\frac{1}{4032},-\frac{19}{1209600}.$$