Let $l=kn-1$. Then we want to find $$R:=\operatorname{Res} \left(\frac{z^l}{ (1-(z/r)^m)^{k}};r\right)=\operatorname{Res}\left(\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k};0\right).$$ This is the coefficient of $z^{-1}$ in the Laurent series expansion of $$\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k}.$$ If we replace $z$ with $rz$, we divide the coefficient of $z^{-1}$ by $r$, so $$R=(-1)^k r^{l+1}[z^{-1}]\left(\frac{(1+z)^l}{\bigl((1+z)^m-1\bigr)^k}\right), $$ where $[z^i]$ extracts the coefficient of $z^i$.

We have $$ \frac{1}{\bigl((1+z)^m-1\bigr)^k}=(mz)^{-k}\left(\frac{mz}{(1+z)^m-1}\right)^k, $$ and since $(1+z)^m -1 = mz\bigl(1+\frac{1}{2}(m-1)z+\frac{1}{6}(m-1)(m-2)z^2+\cdots\bigr)$, we have $$\left(\frac{mz}{(1+z)^m-1}\right)^k=\sum_{i=0}^\infty P_i(m) z^i,$$ where $P_i(m)$ is a polynomial of degree $i$. Thus $$ \begin{aligned} R&=(-1)^k r^{l+1}m^{-k}[z^{k-1}] (1+z)^l \sum_{i=0}^\infty P_i(m) z^i\\ &=(-1)^k r^{l+1}m^{-k} \sum_{j=0}^{k-1} \binom{l}{j} P_{k-1-j}(m)\\ &=(-1)^k r^{kn}m^{-k} \sum_{j=0}^{k-1} \binom{kn-1}{j} P_{k-1-j}(m). \end{aligned} $$ The polynomials $P_i(m)$ are essentially ``higher-order degenerate Bernoulli numbers".

Let $l=kn-1$. Then we want to find $$R:=\operatorname{Res} \left(\frac{z^l}{ (1-(z/r)^m)^{k}};r\right)=\operatorname{Res}\left(\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k};0\right).$$ This is the coefficient of $z^{-1}$ in the Laurent series expansion of $$\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k}.$$ If we replace $z$ with $rz$, we divide the coefficient of $z^{-1}$ by $r$, so $$R=(-1)^k r^{l+1}[z^{-1}]\left(\frac{(1+z)^l}{\bigl((1+z)^m-1\bigr)^k}\right), $$ where $[z^i]$ extracts the coefficient of $z^i$.

We have $$ \frac{1}{\bigl((1+z)^m-1\bigr)^k}=(mz)^{-k}\left(\frac{mz}{(1+z)^m-1}\right)^k, $$ and since $(1+z)^m -1 = mz\bigl(1+\frac{1}{2}(m-1)z+\frac{1}{6}(m-1)(m-2)z^2+\cdots\bigr)$, we have $$\left(\frac{mz}{(1+z)^m-1}\right)^k=\sum_{i=0}^\infty P_i(m) z^i,$$ where $P_i(m)$ is a polynomial of degree $i$. Thus $$ \begin{aligned} R&=(-1)^k r^{l+1}m^{-k}[z^{k-1}] (1+z)^l \sum_{i=0}^\infty P_i(m) z^i\\ &=(-1)^k r^{l+1}m^{-k} \sum_{j=0}^{k-1} \binom{l}{j} P_{k-1-j}(m)\\ &=(-1)^k r^{kn}m^{-k} \sum_{j=0}^{k-1} \binom{kn-1}{j} P_{k-1-j}(m). \end{aligned} $$ The polynomials $P_i(m)$ are essentially "higher-order degenerate Bernoulli numbers".

Let $l=kn-1$. Then we want to find $$R:=\mathrm{Res} \left(\frac{z^l}{ (1-(z/r)^m)^{k}};r\right)=\mathrm{Res}\left(\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k};0\right).$$ This is the coefficient of $z^{-1}$ in the Laurent series expansion of $$\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k}.$$ If we replace $z$ with $rz$, we divide the coefficient of $z^{-1}$ by $r$, so $$R=(-1)^k r^{l+1}[z^{-1}]\left(\frac{(1+z)^l}{\bigl((1+z)^m-1\bigr)^k}\right), $$ where $[z^i]$ extracts the coefficient of $z^i$.

We have $$ \frac{1}{\bigl((1+z)^m-1\bigr)^k}=(mz)^{-k}\left(\frac{mz}{(1+z)^m-1}\right)^k, $$ and since $(1+z)^m -1 = mz\bigl(1+\frac{1}{2}(m-1)z+\frac{1}{6}(m-1)(m-2)z^2+\cdots\bigr)$, we have $$\left(\frac{mz}{(1+z)^m-1}\right)^k=\sum_{i=0}^\infty P_i(m) z^i,$$ where $P_i(m)$ is a polynomial of degree $i$. Thus $$ \begin{aligned} R&=(-1)^k r^{l+1}m^{-k}[z^{k-1}] (1+z)^l \sum_{i=0}^\infty P_i(m) z^i\\ &=(-1)^k r^{l+1}m^{-k} \sum_{j=0}^{k-1} \binom{l}{j} P_{k-1-j}(m)\\ &=(-1)^k r^{kn}m^{-k} \sum_{j=0}^{k-1} \binom{kn-1}{j} P_{k-1-j}(m). \end{aligned} $$ The polynomials $P_i(m)$ are essentially ``higher-order degenerate Bernoulli numbers".

Let $l=kn-1$. Then we want to find $$R:=\operatorname{Res} \left(\frac{z^l}{ (1-(z/r)^m)^{k}};r\right)=\operatorname{Res}\left(\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k};0\right).$$ This is the coefficient of $z^{-1}$ in the Laurent series expansion of $$\frac{(z+r)^l}{\bigl(1-(1+z/r)^m\bigr)^k}.$$ If we replace $z$ with $rz$, we divide the coefficient of $z^{-1}$ by $r$, so $$R=(-1)^k r^{l+1}[z^{-1}]\left(\frac{(1+z)^l}{\bigl((1+z)^m-1\bigr)^k}\right), $$ where $[z^i]$ extracts the coefficient of $z^i$.

We have $$ \frac{1}{\bigl((1+z)^m-1\bigr)^k}=(mz)^{-k}\left(\frac{mz}{(1+z)^m-1}\right)^k, $$ and since $(1+z)^m -1 = mz\bigl(1+\frac{1}{2}(m-1)z+\frac{1}{6}(m-1)(m-2)z^2+\cdots\bigr)$, we have $$\left(\frac{mz}{(1+z)^m-1}\right)^k=\sum_{i=0}^\infty P_i(m) z^i,$$ where $P_i(m)$ is a polynomial of degree $i$. Thus $$ \begin{aligned} R&=(-1)^k r^{l+1}m^{-k}[z^{k-1}] (1+z)^l \sum_{i=0}^\infty P_i(m) z^i\\ &=(-1)^k r^{l+1}m^{-k} \sum_{j=0}^{k-1} \binom{l}{j} P_{k-1-j}(m)\\ &=(-1)^k r^{kn}m^{-k} \sum_{j=0}^{k-1} \binom{kn-1}{j} P_{k-1-j}(m). \end{aligned} $$ The polynomials $P_i(m)$ are essentially ``higher-order degenerate Bernoulli numbers".