Absolute convergence of $\sum\limits_n\left( \log (1-\frac{z}{n})^{n^{k}} + \sum\limits_{m=1}^{k+1}\log e^{n^{k-m}z^m/m} \right)$? 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, 
$$\log\left(\prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\}\right)=$$ $$\bigg\{ \sum_{n=1}^{\infty}n^{k}\log \big(1-\frac{z}{n}\big)\bigg\} + \bigg\{\sum_{m=1}^{k+1} \frac{n^{k-m}z^{m}}{m} \log(e).\bigg\}$$
 A: 
$(1.6)$
$$\sum_{n=1}^{\infty}\log \bigg\{(1-\frac{z}{n})^{n^{k}}\prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}=\bigg\{\sum_{n=1}^{\infty} \log (1-\frac{z}{n})^{n^{k}} + \sum_{m=1}^{k+1}\log e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$

I think this is wrong and the right hand side should be
$$R=\sum_{n=1}^{\infty}\bigg\{ \log (1-\frac{z}{n})^{n^{k}} + \sum_{m=1}^{k+1}\log e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$
From now I suppose plausible claims that $\log$ means $\log_e=\ln$ and $z<1$ in order to take $\log$’s. Then the last expression can be further reduced to
$$\sum_{n=1}^{\infty}\bigg\{ n^{k}\log (1-\frac{z}{n}) + \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg\}.$$
PS. I am sorry, I was suddenly distracted by an UFO. :-) I’ll finish my answer later.
OK, since I was not abducted,  I can continue.
A natural way to prove the absolute convergence of the series for $R$ is the following.
By Taylor's formula in Lagrange form for a function $\log (1+x)$, for each $n$ there exists a number $c_n$ between $0$ and $-\frac zn$ such that
$$\log\left(1-\frac zn\right)=-\sum_{m=1}^{k+1}\frac{z^{m}n^{-m}}{m}+r_n,\mbox{where }r_n=-\frac 1{k+2}\left(\frac {z}{(1+c_n)n}\right)^{k+2}.$$
So $$R=\sum_{n=1}^{\infty} n^kr_n.$$
Then for $z\le 0$ we have
$$| n^kr_n |=\left|\frac 1{(k+2)n^2}\left(\frac {z}{1+c_n}\right)^{k+2}\right|\le \left|\frac {1}{(k+2) n^2}\cdot z^{k+2}\right|.$$
Since the series $\sum_{n=1}^{\infty}\frac {1}{n^2}$ converges (to $\frac{\pi^2}6$), the series for $R$ converges absolutely for each fixed $z\le 0$.
For $0<z<1$ we have
$$| n^kr_n |=\frac 1{(k+2)n^2}\left(\frac {z}{1+c_n}\right)^{k+2}<\frac {z^{k+2}}{(k+2)\left(1-\frac zn\right)^{k+2}n^2}=\frac {1}{(k+2) n^2}\cdot \left(\frac 1z -\frac 1n\right)^{-k-2} .$$
Since the sequence $\left\{\frac 1z -\frac 1n\right\}$ converges to $\frac 1z$, it is bounded, so the series for $R$ converges absolutely for each fixed $0<z<1$.
Remark that for natural $z$ product (1) is zero, because it contains a multiplier $\left(1-\frac{z}{n}\right)^{n^{k}}$, so it diverges.
At last, I remark that if $z>0$ and in product (1) we start to take $\log$’s of multipliers with $n>z$ then we may drop the condition $z<1$, and similarly to the above conclude that the series for $R$ converges absolutely for each fixed $z\in\Bbb R\setminus\Bbb N$.
A: By definition, the infinite product $\prod_{n=1}^\infty (1+a_n)$ converges absolutely iff $\prod_{n=1}^\infty (1+|a_n|)<\infty.$ This happens iff $\sum_{n=1}^{\infty}|a_n|<\infty.$ I'll show your infinite product converges absolutely for all $z\in \mathbb C.$
Define
$$s_n(z) = n^k\left (\frac{z}{n} + \frac{1}{2}\left (\frac{z}{n}\right)^2 + \frac{1}{3}\left (\frac{z}{n}\right)^3 + \cdots + \frac{1}{k+1}\left (\frac{z}{n}\right)^{k+1}\right) .$$
Your product is then
$$\prod_{n=1}^{\infty}(1-z/n)^{n^k}\exp (s_n(z)).$$
Fix $z\in \mathbb C.$ We want to show
$$\tag 1 \sum_{n=1}^{\infty}|(1-z/n)^{n^k}\exp (s_n(z)) - 1| <\infty.$$
Now for $u\in \mathbb C,|u|<1,$ $\log (1-u) = -(u+u^2/2 + u^3/3 + \cdots).$ We can write this as
$$\log(1-u) = -(u+u^2/2 + \cdots + u^{k+1}/(k+1) + r(u)).$$
Let's note that if $|u| \le 1/2,$ then
$$\tag 2 |r(u)| \le \sum_{m=k+2}^\infty |u|^m/m \le \sum_{m=k+2}^\infty |u|^m = |u|^{k+2}(1/(1-|u|) \le 2|u|^{k+2}.$$
For $n$ large,
$$ (1-z/n)^{n^k} = \exp(n^k \log (1-z/n)) = \exp(-s_n(z) + n^kr(z/n)).$$
For such $n,$ the $n$th term in $(1)$ is
$$\tag 3 |\exp(-s_n(z) + n^kr(z/n))\exp (s_n(z)) - 1| = |\exp(n^kr(z/n))-1|.$$
Now for some constant $C,$ $|e^u-1| \le C|u|$ for $|u|<1.$ Thus using $(2),$ and again for large $n,$ $(3)$ is bounded above by
$$ C|n^kr(z/n)| = C|n^k2(z/n)^{k+2}|= 2C|z|^{k+2}\frac{1}{n^2}.$$
Since $\sum_n 1/n^2<\infty,$ we have proved $(1)$ as desired.
