Uniform bound for derivatives of holomorphic function on compact set

I am trying to prove the following statement.

Given $$U \subseteq \mathbb{C}$$ open, a compact set $$K \subset U$$, and $$j \in \mathbb{N}$$, show that there exists a constant $$C > 0$$ such that for any holomorphic $$f : U \to \mathbb{C}$$ and $$z \in K$$ we have $$\begin{equation*} \lvert f^{(j)}(z) \rvert \leq C \sup_{w \in U} \lvert f(w) \rvert. \end{equation*}$$

Here is my attempt. In what follows, $$D(P,r)$$ and $$\overline{D(P, r)}$$ will denote the open and closed disks of radius $$r$$ centered at $$P$$, respectively.

Proof attempt. Let $$z \in K \subset U$$. Since $$U$$ is open, we can find $$r_z > 0$$ such that $$\overline{D(z, r_z)} \subseteq U$$. Cover $$K$$ with $$\{D(z, r_z) \mid z \in K\}$$; since $$K$$ is compact, we can find finitely many $$z_i \in K$$ such that $$\{D(z_i, r_i) \mid 1 \leq i \leq n, r_i := r_{z_i}\}$$ covers $$K$$. Now let $$f : U \to \mathbb{C}$$ be holomorphic and let $$z \in K$$. Then $$z \in D(z_i, r_i)$$ for some $$1 \leq i \leq n$$, so by Cauchy's integral formula and a bound on the path integral we have $$\begin{equation*} \lvert f^{(j)}(z) \vert = \left \vert \frac{j!}{2 \pi i} \oint_{\partial D(z_i, r_i)} \frac{f(w) \ dw}{(w - z)^{j + 1}} \right \vert \leq \frac{j!}{2 \pi} \sup_{w \in \partial D(z_i, r_i)} \left \vert \frac{f(w)}{(w - z)^{j+1}} \right \vert \cdot 2 \pi r_i. \end{equation*}$$ We get $$\begin{equation*} \left \vert w - z \right \vert = \vert (w - z_i) - (z - z_i) \vert \geq \vert \vert w - z_i \vert - \vert z_i - z \vert \vert = \vert r_i - \vert z_i - z \vert \vert \end{equation*}$$
for $$w \in \partial D(z_i, r_i)$$, so $$\begin{equation*} \vert f^{(j)}(z) \vert \leq r_i \cdot j! \sup_{w \in \partial D(z_i, r_i)} \left \vert \frac{f(w)}{(w - z)^{j+1}} \right \vert \leq r_i j! \frac{\sup_{w \in \partial D(z_i, r_i)} \left \vert f(w) \right \vert}{\vert r_i - \vert z_i - z \vert \vert^{j+1}} \leq \frac{j!}{r_i^j} \frac{1}{\left \vert 1 - \frac{\left \vert z_i - z\right \vert}{r_i}\right \vert^{j+1}} \sup_{w \in U} \vert f(w) \vert. \end{equation*}$$

This is as far as I could go. The triangle inequality and the bound I have on $$z_i - z$$ work against me: I can bound it above by $$r_i$$ and below by $$0$$, and both are useless with the reciprocal. If I could get a bound $$M_{i,j}$$ for $$\begin{equation*} \frac{1}{\left \vert 1 - \frac{\left \vert z_i - z \right \vert}{r_i}\right \vert^{j+1}} \end{equation*}$$ that did not depend on $$z$$ then I could take $$C = \max \{M_{i,j} \cdot j!/r_i^j \mid 1 \leq i \leq n\}$$ and the statement would follow. However, I see no way of getting a useful bound.

Any suggestion or hint would be greatly appreciated.

Cover $$K$$ by the disks $$D(z,\frac {r_z} 2)$$ instead of $$D(z,r_z)$$. Then you will have $${|z-z_i|} <\frac {r_i} 2$$ which gives $$|1-\frac {|z-z_i|} {r_i}|^{j+1} \geq (\frac1 2)^{j+1}$$.