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The product converges normally: Pick $K=D(0,r)\subset D(0,1)$ and let $|\cdot|_K=\sup_K |\cdot|$ then
$$\sum_{n=1}^{\infty} |nz^n|_K\leq r|(\frac{1}{1-r}-1)'|=\frac{r}{(1-r)^2}<\infty$$
for each $0<r<1$. This implies that $f$ is holomorphic in $D(0,1)$
Even though my question was had a typo originally, how did you come up with this. I see that it will only converge if |z| < 1 so we need a geometric series but why do you have$$\sum_{n=1}^{\infty} |nz^n|_K\leq r|(\frac{1}{1-r}-1)'|=\frac{r}{(1-r)^2}<\infty$$
I think there's something wrong: Take $z=\frac{1}{2}$ (the estimate works for any positive real number on the unit interval) then for $n>2$ we have $|(1+nz)|^n \geq 1+(nz)^n>2$ which cannot happen if the product converges since individual terms tend to $1$ in that case.
There is a typo sorry. It should say $\prod_{n=1}^{\infty}(1+nz)^n$.
I think there's something wrong: Take $z=\frac{1}{2}$ (the estimate works for any positive real number on the unit interval) then for $n>2$ we have $|(1+nz)|^n \geq 1+(nz)^n>2$ which cannot happen if the product converges since individual terms tend to $1$ in that case.
As for your other question, just notice that $\sum_{n=0}^\infty nz^{n-1}=\left( \frac{1}{1-z}\right)'$ (taking the derivative term by term) and mind the indices.
Ok so I have an idea. $1 - nz^n\Rightarrow z = \sqrt[n]{\dfrac{1}{n}}$. So how do I show that each point on the unit circle is an accumulation point now?How can I show that each point on the unit circle is an accumulation?