$(e^x-1-x)^z$ as a power series

Question

I want to be able to write $$(e^x-1-x)^z$$ as a power series, so since $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ I can write $e^x-1-x$ as $$(e^x-1-x)=\sum_{n=2}^\infty\frac{x^n}{n!}=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}$$ Now, other than trying to expand using brute force, How can I look at the power series expansion of $(e^x-1-x)^z$ for some integer $z$?

Answer 1

Note: As far as I know, there is no closed formula for the coefficient of $x^n$ of $(e^x-1-x)^z$. But, at least you can write the coefficient as sum of multinomial coefficients.

The following notation is convenient. Let $[x^n]$ be the coefficient of operator denoting the coefficient of $x^n$ in a power series $A(x)=\sum_{k=0}^{\infty}a_kx^k$. So, $$[x^n]A(x)=[x^n]\sum_{k=0}^{\infty}a_kx^k=a_n$$

We claim:

The following is valid for $z\in\mathbf{N}$ \begin{align*} [x^n]&(e^x-1-x)^z=\frac{1}{n!}\sum_{{k_1+\dots+k_z=n}\atop{k_1,\dots,k_z\geq 2}}\binom{n}{k_1,\ldots,k_z}\tag{1} \end{align*}

In order to show (1) we calculate:

\begin{align*} [x^n]&(e^x-1-x)^z\\ &=[x^n]\left(\sum_{k=2}^{\infty}\frac{x^k}{k!}\right)^z\\ &=[x^n]\left(\sum_{k_1=2}^{\infty}\frac{x^{k_1}}{k_1!}\right)\cdot\ldots\cdot\left(\sum_{k_z=2}^{\infty}\frac{x^{k_z}}{k_z!}\right)\\ &=[x^n]\sum_{n=2z}^{\infty}\left(\sum_{{k_1+\dots+k_z=n}\atop{k_1,\dots,k_z\geq 2}}\frac{x^{k_1+\dots+k_z}}{k_1!\cdot\ldots\cdot k_z!}\right)\\ &=[x^n]\sum_{n=2z}^{\infty}\left(\sum_{{k_1+\dots+k_z=n}\atop{k_1,\dots,k_z\geq 2}}\frac{1}{k_1!\cdot\ldots\cdot k_z!}\right)x^{n}\\ &=[x^n]\sum_{n=2z}^{\infty}\left(\sum_{{k_1+\dots+k_z=n}\atop{k_1,\dots,k_z\geq 2}}\frac{n!}{k_1!\cdot\ldots\cdot k_z!}\right)\frac{x^{n}}{n!}\\ &=[x^n]\sum_{n=2z}^{\infty}\left(\sum_{{k_1+\dots+k_z=n}\atop{k_1,\dots,k_z\geq 2}}\binom{n}{k_1,\ldots,k_z}\right)\frac{x^{n}}{n!}\\ &=\frac{1}{n!}\sum_{{k_1+\dots+k_z=n}\atop{k_1,\dots,k_z\geq 2}}\binom{n}{k_1,\ldots,k_z} \end{align*}