To prove that $f(z)=z^n$, $n=1,2,3,\dots$ is continuous in $\mathbb{C}$, I know that we need:
Given $\varepsilon>0,\, \exists\, \delta>0$ such that if $\left|z-z_0\right|<\delta$, then $\left|z^n-z_0^n\right|<\varepsilon$.
I'm confused as to where to go from here.
I've tried using the triangle inequality to show
$$\left|z^n-z_0^n\right|\leq\left|z^n\right|-\left|z_0^n\right|=\left|z\right|^n-\left|z_0\right|^n=(x^2+y^2)^{\frac{n}{2}}-(x_0^2+y_0^2)^{\frac{n}{2}}$$
but I'm 1) not sure if that's totally correct and 2) if it is, I'm not sure how to use it to choose $\delta$.
$|z^n-z_{0}^n|=|(z-z_0)(z^{n-1}+z^{n-2}z_0+.....+z_0^{n-1})| \leq |z-z_0||z^{n-1}+z^{n-2}z_0+.....+z_0^{n-1}| $
As, $|z-z_0|<\delta \implies |z|=|z-z_0+z_0|< |z-z_0|+|z_0|=\delta+|z_0|$
$|z^n-z_{0}^n| \leq |z-z_0|(|z|^{n-1}+|z|^{n-2}|z_0|+.....+|z_0|^{n-1})$
$|z^n-z_{0}^n| \leq \delta((\delta+|z_0|)^{n-1}+(\delta+|z_0|)^{n-2}|z_0|+.....+|z_0|^{n-1}) $
You can choose $\delta$ accordingly to get $|z^n-z_{0}^n|< \epsilon$
However, you can also say that $z^n$ is continuous by the fact that it is the product of continuous function $z$ (n times)(Using Induction).