Prove that $\ \prod_{n=1}^{\infty} \{(1-\frac{z}{n})^{nk} \exp(\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}) \}$ converges absolutely.

Question

Prove that $\forall z,\ \prod_{n=1}^{\infty} \{(1-\frac{z}{n})^{nk} \exp(\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}) \}$ where $k$ is a positive integer, converges absolutely.

It seems to me $(1-\frac{z}{n})^{nk}\rightarrow e^{-kz}$, and so the infinite product approximates $$\prod_{n=1}^{\infty} \left\{\exp(-kz) \exp(\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}) \right\}=\prod_{n=1}^{\infty}e^{(n^{k-1}-k)z+\dots},$$

which when $k > 1$ diverges for $z\in\{z, |z|>1\}$, by the ratio test of $\frac{|((n+1)^{k-1}-k)z|}{|(n^{k-1}-k)z|}.$ [edit: this is incorrect, since this power of z (as well as all other powers of z, i.e. m, which depends only on k) is independent of the 'independent variable of the series' n.]

which we from the ratio test we see equals an infinite series $\sum u_n$ with $\frac{|u_{n+1}|} {|u_n|}\rightarrow 1.$

Then by expanding $\frac{|u_{n+1}|} {|u_n|}$ as 1+A/n+O(1/n$^2$), we have A=k-1>-1, ($\sum$ n$^{k-1}$ converges not,) so the infinite product doesn't converge.

My question:

1. is the step 'the infinite product approximates $\prod_{n=1}^{\infty} \{\exp(-kz) \exp(\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}) \}$' valid?
2. Is my conclusion about convergence or not correct? I guess not. If so, where does the proof go wrong?

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Edited to add:

As pointed out in the answer, the original problem is about $$\prod_{n=1}^{\infty} \{(1-\frac{z}{n})^{n^k} \exp(\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}) \}.$$

Answer 1

Replacing $(1-z/n)^{n\color{gray}{k}}$ with $e^{-\color{gray}{k}z}$ is not valid, since the difference of logarithms of these is $k$ times $$n\log\left(1-\frac{z}{n}\right)+z=-\frac{z^2}{2n}+\mathcal{O}(n^{-2})$$ as $n\to\infty$ with $z$ fixed, thus it introduces a divergence of harmonic type $\color{LightGray}{\left[\sum_n 1/n\right]}$ (if $z\neq 0$).

The original claim is wrong as well. Most probably, there's a typo; a fixed version would ask for $$\prod_{k=1}^\infty\left\{\left(1-\frac{z}{n}\right)^{\color{red}{n^k}}\exp\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}\right\}.$$

In all the cases, the approach is to take logs and taylor-expand. For a fixed $z$, and $n$ large enough, $$n^k\log\left(1-\frac{z}{n}\right)+\sum_{m=1}^{k+1}\frac{n^{k-m}z^m}{m}=-\sum_{m=k+2}^\infty\frac{n^{k-m}z^m}{m}=\mathcal{O}(n^{-2}).$$