Coefficient extraction I want to show:
\begin{equation*}
  [z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\alpha + k}}
\end{equation}
So far I got
\begin{equation*}
    \frac{1}{(1 - z)^{\alpha + 1}} = (1-z)^{-(\alpha + 1)} = \sum_{n \geq 0}{\binom{-\alpha - 1}{k}(-1)^k z^k} =  \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n}
  \end{equation*}
where I used $\binom{-\alpha}{k} = (-1)^k \binom{\alpha + k - 1}{k}$and
\begin{equation*}
    \log \frac{1}{1 - z} = - \log 1- z = \sum_{n \geq 1}\frac{z^n}{n} = z\sum_{n \geq 0}\frac{z^{n}}{n+1}
  \end{equation*}
therefore
\begin{align*}
    \frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} &= z\left( \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n} \right) \left( \sum_{n \geq 0}\frac{z^{n}}{n+1} \right) \\
    &= z \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n} \\
    &= \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n + 1}.
  \end{align*}
Now the $n$-th coefficient is
\begin{align*}
    \sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\frac{n!}{(n-k)! k!} \frac{1}{\binom{\alpha + n}{n - k}}    \frac{1}{n - k + 1}} +  \frac{1}{n+1}\\
    &=  \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k} \frac{1}{\binom{\alpha + n}{n - k}}    \frac{1}{n - k + 1}}  +   \frac{1}{n+1}\\
    &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k - 1} \frac{1}{\binom{\alpha + n}{n - k - 1}} \frac{k-1}{n - k -1}  \frac{1}{\alpha + k + 1} } + \frac{1}{n+1}\\
    &= \binom{\alpha + n}{n} \sum^{n+1}_{k = 2}{\binom{n}{k - 2} \frac{1}{\binom{\alpha + n}{n - k}} \frac{k- 2 }{n - k}  \frac{1}{\alpha + k} } + \frac{1}{n+1}
  \end{align*}
but from now on I can't see how to proceed.
I also know that
\begin{equation}
[z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n
\end{equation}
somehow I also tried to use some sort of transformation law for
coefficient extraction, but I am not aware of any kind of transformation law for coefficient extraction. Using $1-u = (1-z)^{\alpha + 1}$ or $z = 1 - (1-u)^{1/(\alpha + 1)}$gives
\begin{equation}
\frac{1}{\alpha + 1}\frac{1}{1-u} \log \frac{1}{1 - u}
\end{equation}
and for this I get the coefficient $\frac{1}{\alpha + 1} H_n$.
Another approach I did was the following:
Since
\begin{equation*}
    (1-z)^{-m} = \exp(-m \ln(1-z)) 
  \end{equation*}
I get
\begin{equation*}
    \frac{\partial }{\partial m} \exp(-m \ln(1-z))  = -\ln(1-z)  \exp(- m \ln(1-z)) = - \ln(1-z) \frac{1}{(1-z)^m} = \frac{1}{(1-z)^m} \ln \frac{1}{1-z}
  \end{equation*}
therefore
\begin{align*}
    \frac{\partial }{\partial m} (1-z)^{-m} &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \binom{m + n - 1}{n} z^n} \\
    &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \frac{\Gamma(m + n)}{n!\Gamma(m)} z^n} 
\end{align*}
But I don't know any identities for the derivative of the gamma function.
 A: We will need an auxiliary identity before we can move on to the subject
by OP, which is
$$[z^n] \frac{1}{(1-z)^{\alpha+1}} \log\frac{1}{1-z}
= {n+\alpha\choose n} (H_{n+\alpha} - H_\alpha).$$
with $\alpha$ a non-negative integer.
 A binomial identity 
Introduce with $q\ge 1$
$$f(z) = n! (-1)^n \frac{1}{z+q} \prod_{p=0}^n \frac{1}{z-p}.$$
This has the property that for $0\le r\le n$
$$\mathrm{Res}_{z=r} f(z)
= n! (-1)^n \frac{1}{r+q}
\prod_{p=0}^{r-1} \frac{1}{r-p}
\prod_{p=r+1}^n \frac{1}{r-p}
\\ = n! (-1)^n \frac{1}{r+q}
\frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!}
= {n\choose r} \frac{(-1)^r}{r+q}.$$
With the residue at infinity being zero by inspection we obtain
$$\sum_{r=0}^n {n\choose r} \frac{(-1)^r}{r+q}
= - \mathrm{Res}_{z=-q} f(z)
\\ = - n! (-1)^n \prod_{p=0}^n \frac{1}{-q-p}
= n! \prod_{p=0}^n \frac{1}{q+p}
= n! \frac{(q-1)!}{(q+n)!}
= \frac{1}{q} {n+q\choose q}^{-1}.$$
Therefore with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= \sum_{r=0}^{n-k} {n-k\choose r} \frac{(-1)^r}{r+k}
= (-1)^{n-k} \sum_{r=0}^{n-k} {n-k\choose r} \frac{(-1)^r}{n-r}
\\ = [z^n] \log\frac{1}{1-z}
(-1)^{n-k} \sum_{r=0}^{n-k} {n-k\choose r} (-1)^r z^r
\\ = [z^n] \log\frac{1}{1-z}
(-1)^{n-k} (1-z)^{n-k}.$$
 Main identity 
We get for the LHS from first principles that it is (apply identity
setting $n$ to $n+\alpha$ and $k$ to $q$)
$$\sum_{q=1}^n {n-q+\alpha\choose n-q} \frac{1}{q}
\\ = [z^{n+\alpha}] \log\frac{1}{1-z}
\sum_{q=1}^n {n+\alpha\choose q} {n-q+\alpha\choose \alpha}
(-1)^{n+\alpha-q} (1-z)^{n+\alpha-q}.$$
Note that for $q=0$ we get
$${n+\alpha\choose \alpha} 
[z^{n+\alpha}] \log\frac{1}{1-z} 
(-1)^{n+\alpha} (1-z)^{n+\alpha}.$$
This will be our first piece. We include it in our sum at this time. Next
observe that
$${n+\alpha\choose q} {n-q+\alpha\choose \alpha}
= \frac{(n+\alpha)!}{q! \times \alpha! \times (n-q)!}
= {n+\alpha\choose \alpha} {n\choose q}.$$
We have for the augmented sum without the binomial scalar in front
$$[z^{n+\alpha}] \log\frac{1}{1-z}
\sum_{q=0}^n {n\choose q} (z-1)^{n+\alpha-q}
\\ = [z^{n+\alpha}] \log\frac{1}{1-z}
(z-1)^{n+\alpha} \left[1+\frac{1}{z-1}\right]^n
\\ = [z^{n+\alpha}] \log\frac{1}{1-z}
(z-1)^\alpha z^n
= [z^\alpha] \log\frac{1}{1-z} (z-1)^\alpha.$$
This is the second piece. Now to evaluate these two pieces
we evidently require
$$\;\underset{z}{\mathrm{res}}\;
\frac{1}{z^{m+1}} \log\frac{1}{1-z} (-1)^m (1-z)^m.$$
We put $z/(1-z) = w$ so that $z=w/(1+w)$ and $dz = 1/(1+w)^2 \; dw$ to
obtain
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{m+1}} (1+w)
\log\frac{1}{1-w/(1+w)} (-1)^m \frac{1}{(1+w)^2}
\\ = (-1)^m \;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{m+1}} \frac{1}{1+w} \log(1+w)
= (-1)^m [w^m] \frac{1}{1+w} \log(1+w)
\\ = [w^m] \frac{1}{1-w} \log(1-w) 
= - [w^m] \frac{1}{1-w} \log\frac{1}{1-w} 
= - H_m.$$
Hence our first piece is $- {n+\alpha\choose\alpha} H_{n+\alpha}$ while
the second is $- {n+\alpha\choose\alpha} H_{\alpha}$. Subtract the
first from the second to obtain our claim,
$${n+\alpha\choose n} (H_{n+\alpha} - H_\alpha).$$
A: Here is another variation. We show the identity
\begin{align*}
\color{blue}{[z^n]\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{1}
\end{align*}
is valid for integral $n>0$ and $\alpha \in\mathbb{C}\setminus\{-1,\ldots,-n\}$.

We start with the LHS of (1) and obtain
\begin{align*}
\color{blue}{[z^n]}\color{blue}{\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}}
&=[z^n]\frac{1}{(1-z)^{\alpha+1}}\sum_{k=1}^\infty\frac{z^k}{k}\tag{2.1}\\
&=\sum_{k=1}^n\frac{1}{k}[z^{n-k}]\frac{1}{(1-z)^{\alpha+1}}\tag{2.2}\\
&=\sum_{k=1}^n\frac{1}{k}\binom{-\alpha-1}{n-k}(-1)^{n-k}\tag{2.3}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}}\tag{2.4}
\end{align*}

Comment:

*

*In (2.1) we use the logarithmic series expansion.


*In (2.2) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and restrict the upper limit of the sum to $n$, since other terms do not contribute.


*In (2.3) we select the coefficient of $[z^{n-k}]$ of the binomial series expansion.


*In (2.4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

It remains to show the binomial identity
\begin{align*}
\color{blue}{\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{3}
\end{align*}

A nice aspect is, that when considering the expressions in (3) as polynomial in $\alpha$ we can show, that they both are the derivative $\frac{\partial}{\partial\alpha}\binom{n+\alpha}{n}$.

We start with the RHS of (3) which is the easy part. We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial \alpha}}\color{blue}{\binom{n+\alpha}{n}}
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=0}^{n-1}(n+\alpha-j)\tag{4.1}\\
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=0}^{n-1}(\alpha+j+1)\tag{4.2}\\
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=1}^{n}(\alpha+j)\tag{4.3}\\
&=\frac{1}{n!}\sum_{k=1}^n\left(\frac{\partial}{\partial \alpha}(\alpha+k)\right)\prod_{{j=1}\atop {j\ne k}}^n(\alpha+j)\tag{4.4}\\
&=\frac{1}{n!}\sum_{k=1}^n\frac{1}{\alpha+k}\prod_{j=1}^n(\alpha+j)\tag{4.5}\\
&\,\,\color{blue}{=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}
\end{align*}
and we get the RHS of (3).

Comment:

*

*In (4.1) we use the definition of binomial coefficients in the form $\binom{p}{q}=\frac{1}{q!}\prod_{j=0}^{q-1}(p-j)$.


*In (4.2) we change the order of multiplication $j\to n-1-j$.


*In (4.3) we shift the index and start with $j=1$.


*In (4.4) we use the product rule for derivation of products with $n$ terms.


*In (4.5) we do the derivation and expand numerator and denominator with $\alpha+k$.

Now the somewhat more challenging part. We do the derivation by using first principles. We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial \alpha}}&\color{blue}{\binom{n+\alpha}{n}}
=(-1)^n\frac{\partial}{\partial \alpha}\binom{-\alpha-1}{n}\tag{5.1}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\left(\binom{-\alpha-1-q}{n}-\binom{-\alpha-1}{n}\right)\tag{5.2}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\left(\sum_{k=0}^n\binom{-q}{k}\binom{-\alpha-1}{n-k}-\binom{-\alpha-1}{n}\right)\tag{5.3}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\sum_{k=1}^n\binom{-q}{k}\binom{-\alpha-1}{n-k}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\sum_{k=1}^n\frac{-q}{k}\binom{-q-1}{k-1}\binom{-\alpha-1}{n-k}\tag{5.4}\\
&=(-1)^{n+1}\sum_{k=1}^n\frac{1}{k}\binom{-1}{k-1}\binom{-\alpha-1}{n-k}\\
&=\sum_{k=1}^n\frac{(-1)^{n-k}}{k}\binom{-\alpha-1}{n-k}\tag{5.5}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}}\tag{5.6}
\end{align*}
which is the LHS of (3) and the claim (3) follows.

Comment:

*

*In (5.1) we use again the binomial identity as in (2.4).


*In (5.2) we do the derivation according to first principles.


*In (5.3) we apply the Chu-Vandermonde identity.


*In (5.4) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (5.5) we use the binomial identity $\binom{-1}{q}=\frac{1}{q!}(-1)(-2)\cdots(-q)=(-1)^q$.


*In (5.6) we use again the binomial identity as in (2.4).
A: Perhaps similar to epi163sqrt's answer in that this uses Vandermonde's Identity. However, this avoids the use of a derivative.

Using the power series
$$
\log\left(\frac1{1-z}\right)=\sum_{k=1}^\infty\frac{z^k}{k}\tag1
$$
and the Generalized Binomial Theorem with negative binomial coefficents
$$
\begin{align}
(1-z)^{-\alpha-1}
&=\sum_{k=0}^\infty\binom{-\alpha-1}{k}(-1)^kz^k\tag{2a}\\
&=\sum_{k=0}^\infty\binom{k+\alpha}{k}z^k\tag{2b}
\end{align}
$$
we get
$$
\begin{align}
&\left[z^n\right] (1-z)^{-\alpha-1}\log\left(\frac1{1-z}\right)\tag{3a}\\
&=\sum_{k=0}^{n-1}\frac1{n-k}\binom{k+\alpha}{k}\tag{3b}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{\binom{k+\alpha}{k}}{\binom{n+\alpha}{n}}\tag{3c}\\
%&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{\frac{(k+\alpha)\dots(1+\alpha)}{k!}}{\frac{(n+\alpha)\dots(1+\alpha)}{n!}}\tag{3d}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{n\dots(k+1)}{(n+\alpha)\dots(k+1+\alpha)}\tag{3d}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac1{n-k}\frac{\binom{n}{n-k}(n-k)!}{(n+\alpha)\dots(k+1+\alpha)}\tag{3e}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\frac{\binom{n}{n-k}(n-k-1)!}{(n+\alpha)\dots(k+1+\alpha)}\tag{3f}\\
&=\binom{n+\alpha}{n}\sum_{k=0}^{n-1}\sum_{j=k+1}^n\frac1{j+\alpha}\frac{\binom{n}{n-k}(n-k-1)!}{(n-j)\dots1(-1)(k+1-j)}\tag{3g}\\
&=\binom{n+\alpha}{n}\sum_{j=1}^n\sum_{k=0}^{j-1}\frac1{j+\alpha}(-1)^{j-k-1}\binom{n}{n-k}\binom{n-k-1}{j-k-1}\tag{3h}\\
&=\binom{n+\alpha}{n}\sum_{j=1}^n\sum_{k=0}^{j-1}\frac1{j+\alpha}\binom{n}{k}\binom{j-n-1}{j-k-1}\tag{3i}\\
&=\binom{n+\alpha}{n}\sum_{j=1}^n\frac1{j+\alpha}\tag{3j}\\
&=\binom{n+\alpha}{n}\left(H_{n+\alpha}-H_\alpha\right)\tag{3k}
\end{align}
$$
Explanation:
$\text{(3b)}$: $(1)$, $(2)$, and the Cauchy Product Formula
$\text{(3c)}$: factor $\binom{n+\alpha}{n}$ out front
$\text{(3d)}$: expand and cancel the binomial coefficients
$\text{(3e)}$: rewrite the product in the numerator
$\text{(3f)}$: cancel the $n-k$ in the numerator and denominator
$\text{(3g)}$: apply partial fractions
$\text{(3h)}$: collect the factors into $(-1)^{j-k-1}\binom{n-k-1}{j-k-1}$
$\text{(3i)}$: apply negative binomial coefficients
$\text{(3j)}$: apply Vandermonde's Identity
$\text{(3k)}$: rewrite the sum as a difference of Extended Harmonic Numbers
