I'm having trouble verifying my appoarch to the problem in $(1)$, much of efforts can be seen in the sections titled $\text{Lemma}$, I'm specifically stuck where $(1.6)$ would the absolute convergence of the double series imply anything about the absolute convergence of the product in $(1)$ ?
$(1)$
Prove that
$$\prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\}$$
where $k$ is any positive integer, converges absolutely for all values of $z.$
$\text{Lemma (1.2)}$
On the RHS side of $(1)$, one can note the following in $(1.3)$
$(1.3)$
$$\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) = \prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \text{.}$$
$\text{Remark}$
The developments in $(1.3)$ is due to the notion that one can go from sum to product, where:
$$\exp \sum s_n = \prod e^{s_n}$$
Using what we've discovered in $(1.3)$, we can make the following deductions in $(1.5)$
$(1.5)$
$$\log \prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\} = \sum_{n=1}^{\infty}\log \bigg\{(1-\frac{z}{n})^{n^{k}}\prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$
$\text{Lemma (1.3)}$
Distributing the $\log$, on the RHS side of $(1.5)$, we obtain a product of two series in $(1.6)$
$(1.6)$
$$\sum_{n=1}^{\infty}\log \bigg\{(1-\frac{z}{n})^{n^{k}}\prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}=\bigg\{\sum_{n=1}^{\infty} \log (1-\frac{z}{n})^{n^{k}} + \sum_{m=1}^{k+1}\log e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$
Finally, condensing our coefficients we have in $(1.7)$
$(1.7)$
$$\sum_{n=1}^{\infty}n^{k}\log \big(1-\frac{z}{n} \big) + \sum_{m=1}^{k+1} \frac{n^{k-m}z^{m}}{m} \log(e).$$
In summary,
$$\log\left(\prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\}\right)=$$ $$\bigg\{ \sum_{n=1}^{\infty}n^{k}\log \big(1-\frac{z}{n}\big)\bigg\} + \bigg\{\sum_{m=1}^{k+1} \frac{n^{k-m}z^{m}}{m} \log(e).\bigg\}$$