Proving uniform convergence on disk within radius of convergence Needham's Visual Complex Analysis 2.III.2 states that a power series $S_k=\sum{C_k z^k}$ with RoC $R$ converges uniformly on any disk $r<R$.
He leaves the proof as an exercise to the reader. But this reader is struggling!
The only approach I can think of is to try supposing that uniform convergence fails on the disk, and argue for a contradiction.
Convergence is guaranteed everywhere on the disk, as $r < R$. Choosing $\epsilon>0$, each point $z$ on the disk will have an associated N s.t. $S_{n>N}(z)$ is within $\epsilon$ of $S_\infty(z)$.
So maybe we could consider the supremum of these N, and argue against it being infinite...?  But this approach looks against the character of the book. I can't believe Needham has this in mind.
I suspect there is some visual way to see it.
But how?
 A: Hint: if the series converges at a point with $|z|=r_1$, its terms are bounded there.  Use that to get an exponentially decreasing uniform bound on $|z| \le r$, where $r < r_1 < R$.
A: Here's a few more details. Let $r < R$, where $R$ is the radius of convergence of
$$
\sum_{k=0}^\infty c_k z^k.
$$
Fix a point $p$ with $|p|$ a little larger than $r$ and a little smaller than $R$, for example, $|p| = (r+R)/2$. Since the series converges for $p$, we know that $c_k p^k \to 0$ as $k\to \infty$, and so does $|c_k p^k|$. 
To show uniform convergence on $|z|\le r$, we get
$$
M_k = \sup_{|z| \le r} |c_kz^k| \le |c_k| r^k = |c_k p^k| \cdot \frac{r^k}{|p|^k}
$$
Since  $|c_k p^k| \to 0$, there is a constant $A$ (independent of $k$) such that  $|c_k p^k| \le A$ for all $k$, i.e. $M_k \le A (r/|p|)^k$. Hence
$$
\sum_{k=1}^\infty M_k
$$
converges (by comparison with a convergent geometric series; the ratio is $r/|p| < 1$), and Weierstrass' $M$-test gives us uniform convergence.
A: mrf has provided the solution (thanks!). Nevertheless I will answer in my own words to make sure my understanding is clear:
We have $S_N(z) = \sum^N C_k z^k$ with RoC $R$, and wish to demonstrate uniform convergence on a disc radius $r < R$.
Consider all $\text{N-tail}(z) = C_{N+1} z^{N+1} + C_{N+2}z^{N+2} + ...$ for $z$ on/in r-disc. Each tail is the difference between $S_{N}(z)$ and $S_\infty(z)$ i.e. the error of $S_N(z)$.
We want to show that for a given N, we can find a bound B(N) for all N-tails (errors) on the r-disc, and $B(N) \to 0$ as $N \to \infty$
Take a (real) point $p$ midway between $r$ and $R$. $\sum C_k p^k$ converges, therefore $|C_k|p^k \to 0$, therefore $|C_k|p^k < M$ or $|C_k| < \frac{M}{p^k}$.
We now have a bound for $|C_k|$ in terms of $k$.
We can also observe that if $ |z| \leq r $ then $ |z^k| \leq r^k $
We can use the above to place a bound on $\text{N-tail}(z)$ for $|z| \leq r$  that does not depend upon z, only N:
$\hspace{1cm} | \text{N-tail}(z) | 
= |C_{N+1} z^{N+1} + C_{N+2}z^{N+2} + ...|
\leq \sum \limits_{N+1}^\infty |C_k| |z^k| 
\leq \sum \limits_{N+1}^\infty \frac{M}{p^k} r^k \\
\hspace{2cm} = M \sum \limits_{N+1}^\infty \rho^k \text{ where }\rho = \frac{r}{p} \in (0,1) \\
\hspace{2cm} = M \rho^{N+1} (1+\rho+\rho^2+...) = \frac{M}{1-\rho}\rho^{N+1}$
As $|\rho|< 1$ this is clearly dwindles to 0 as $N \to \infty$
