The $n$-th derivative of the reciprocal of a function and a binomial identity

While I was looking for an answer to this MSE post in order to prove
\begin{align*}
\frac{d^n}{dx^n}\left(\frac{1}{1-e^{-x}}\right)=(-1)^n\sum_{j=1}^n{n\brace j}j!\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\tag{1}
\end{align*}
with the numbers ${n\brace j}$ denoting the Stirling numbers of the second kind I considered the following formula of the reciprocal of the $n$-th derivative of a function which might be interesting by itself.

With $D_x:=\frac{d}{dx}$ the following relationship is valid according to (3.63) in H.W. Goulds Binomial Identities, vol. I

\begin{align*}
D_x^n\left(\frac{1}{f(x)}\right)=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{1}{\left(f(x)\right)^{j+1}}D_x^n\left(\left(f(x)\right)^j\right)
\end{align*}

$$ $$

Applying this formula to the function $f(x)=\frac{1}{1-e^{-x}}$ we obtain
\begin{align*}
D_x^n&\left(\frac{1}{1-e^{-x}}\right)\\
&=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{1}{\left(1-e^{-x}\right)^{j+1}}D_x^n\left(\left(1-e^{-x}\right)^j\right)\\
&=\sum_{j=0}^n\frac{(-1)^j}{\left(1-e^{-x}\right)^{j+1}}\binom{n+1}{j+1}D_x^n\left(\sum_{k=0}^j\binom{j}{k}(-1)^ke^{-kx}\right)\\
&=(-1)^n\sum_{j=0}^n\frac{(-1)^j}{\left(1-e^{-x}\right)^{j+1}}\binom{n+1}{j+1}\sum_{k=0}^j\binom{j}{k}(-1)^kk^ne^{-kx}\\
&=\frac{(-1)^n}{(1-e^{-x})^{n+1}}\sum_{j=1}^n(-1)^j\left(1-e^{-x}\right)^{n-j}\binom{n+1}{j+1}\sum_{k=1}^j\binom{j}{k}(-1)^kk^ne^{-kx}\tag{2}
\end{align*}

On the other hand with the identity ${n\brace j}=\frac{1}{j!}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^n$

we obtain from (1)
\begin{align*}
D_x^n&\left(\frac{1}{1-e^{-x}}\right)\\
&=(-1)^n\sum_{j=1}^n{n\brace j}j!\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\\
&=(-1)^n\sum_{j=1}^n\frac{e^{-jx}}{\left(1-e^{-jx}\right)^{j+1}}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^n\\
&=\frac{(-1)^n}{(1-e^{-x})^{n+1}}\sum_{j=1}^n(-1)^je^{-jx}(1-e^{-x})^{n-j}\sum_{k=1}^j\binom{j}{k}(-1)^{k}k^n\tag{3}
\end{align*}

I have difficulties to prove the equality of (2) with (3). So putting $y=e^{-x}$ I would like to ask for a prove of the following relationship

Claim: The following is valid for $n\geq 1$ and $y\geq 0$.
\begin{align*}
\sum_{j=1}^n&(-1)^j\left(1-y\right)^{n-j}\binom{n+1}{j+1}\sum_{k=1}^j\binom{j}{k}(-1)^kk^ny^k\\
&=\sum_{j=1}^n(-1)^jy^j(1-y)^{n-j}\sum_{k=1}^j\binom{j}{k}(-1)^{k}k^n
\end{align*}

Please note I’m not interested in a proof by induction. I would like to see how to transform one side into the other, maybe with the help of generating functions.

Many thanks in advance.

Solutions Collecting From Web of "The $n$-th derivative of the reciprocal of a function and a binomial identity"

Extracting coefficients on $[y^m]$ where $0\le m\le n$ we see that we
have to prove that

$$\sum_{j=1}^n (-1)^j {n+1\choose j+1}
\sum_{k=1}^j {j\choose k} (-1)^k k^n {n-j\choose m-k} (-1)^{m-k}
\\ = \sum_{j=1}^n (-1)^j {n-j\choose m-j} (-1)^{m-j}
\sum_{k=1}^n (-1)^{k} {j\choose k} k^n$$

or alternatively

$$\sum_{j=1}^n (-1)^j {n+1\choose j+1}
\sum_{k=1}^j {j\choose k} k^n {n-j\choose m-k}
= \sum_{j=1}^n {n-j\choose m-j}
\sum_{k=1}^n (-1)^{k} {j\choose k} k^n.$$

Re-write this as follows:

$$\sum_{k=1}^n k^n \sum_{j=k}^n
(-1)^j {n+1\choose j+1} {j\choose k} {n-j\choose m-k}
= \sum_{k=1}^n k^n (-1)^k
\sum_{j=1}^n {n-j\choose m-j} {j\choose k}.$$

We have the claim if we can show that

$$ \sum_{j=k}^n
(-1)^j {n+1\choose j+1} {j\choose k} {n-j\choose m-k}
= (-1)^k \sum_{j=1}^n {n-j\choose m-j} {j\choose k}.$$

For the LHS we introduce

$${j\choose k} = {j\choose j-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{j-k+1}} (1+z)^j \; dz$$

This has the property that it vanishes when $j\lt k$ including $j=-1$
so we may lower the index of the sum to $-1.$ We also introduce

$${n-j\choose m-k}
= \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n-j} \; dw$$

and obtain

$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
z^{k-1}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n}
\\ \times \sum_{j=-1}^n {n+1\choose j+1} (-1)^j
\frac{(1+z)^j}{z^j (1+w)^j}
\; dw\; dz
\\ = – \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^{k}}{1+z}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n+1}
\\ \times \sum_{j=-1}^n {n+1\choose j+1} (-1)^{j+1}
\frac{(1+z)^{j+1}}{z^{j+1} (1+w)^{j+1}}
\; dw\; dz
\\ = – \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^{k}}{1+z}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(1+w)^{n+1}
\\ \times \left(1-\frac{1+z}{z(1+w)}\right)^{n+1}
\; dw\; dz
\\ = – \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \frac{1}{1+z}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{m-k+1}}
(wz-1)^{n+1}
\; dw\; dz.$$

Extracting coefficients we find

$$- (-1)^{n+1-m+k} {n+1\choose m-k}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} \frac{1}{1+z}
z^{m-k} \; dz
\\ = (-1)^k {n+1\choose m-k}.$$

Observe that we didn’t use the differential in the integral which
means this also goes through using formal power series
only. Continuing with the RHS we find

$$(-1)^k \sum_{j=k}^n {n-j\choose m-j} {j\choose k}
= (-1)^k \sum_{j=0}^{n-k} {n-k-j\choose m-k-j} {j+k\choose k}.$$

We introduce

$${n-k-j\choose m-k-j}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k-j+1}} (1+z)^{n-k-j} \; dz$$

This vanishes when $j\gt m-k$ at some point at most at the upper index
(recall that $m\le n$). Hence we are justified in extending $j$ to
infinity and obtain

$$\frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k+1}} (1+z)^{n-k}
\sum_{j\ge 0} {j+k\choose k} \frac{z^j}{(1+z)^j}
\; dz
\\ = \frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k+1}} (1+z)^{n-k}
\frac{1}{(1-z/(1+z))^{k+1}}
\; dz
\\ = \frac{(-1)^k}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-k+1}} (1+z)^{n+1}
\frac{1}{(1+z-z)^{k+1}}
\; dz
= (-1)^k {n+1\choose m-k}.$$

We have shown that the coefficients on $k^n$ in the outer sum are
equal and hence so are the coefficients on $[y^m].$ This concludes the
argument.