= Solution
The <Cauchy integral formula> states that if $f$ is <holomorphic> on a neighbourhood of a simple closed positively oriented contour and its interior, then, for an interior point $z$,
$$
f(z)=\frac1{2\pi i}\oint\frac{f(\xi)}{\xi-z}\,d\xi.
$$
Fix $0<r<1$ and apply it on $|\xi|=r$. For $|z|<r$, the <geometric series>
$$
\frac1{\xi-z}=\sum_{n=0}^{\infty}\frac{z^n}{\xi^{n+1}}
$$
converges uniformly in $\xi$ and locally uniformly in $z$. Indeed, for $|z|\leq s<r$ its absolute terms are bounded by $s^n/r^{n+1}$. Since $f$ is bounded on the contour, termwise <contour integration> is valid, giving
$$
f(z)=\sum_{n=0}^{\infty}c_nz^n,\qquad
c_n=\frac1{2\pi i}\oint_{|\xi|=r}\frac{f(\xi)}{\xi^{n+1}}\,d\xi.
$$
Termwise differentiation of this locally convergent <power series> gives $f^{(n)}(0)=n!c_n$. Therefore
$$
\boxed{f^{(n)}(0)=\frac{n!}{2\pi i}\oint_{|\xi|=r}\frac{f(\xi)}{\xi^{n+1}}\,d\xi\quad(n\geq0)}.
$$
This also derives the needed <Taylor series> representation rather than assuming an unjustified interchange at the boundary of the analytic disk.
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