<s>[INST] 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.
 [/INST] 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
</s>