 Solution

The Cauchy derivative formula says that if $f$ is holomorphic on a neighbourhood of a closed disk and $z_0$ is inside its positively oriented boundary circle $\gamma$, then for $n\ge 0$,For $n=0$ this is the Cauchy integral formula. The same formula holds for a positively oriented simple closed contour when $f$ is holomorphic on a neighbourhood of the contour and its enclosed region.

In the present problem the unit circle and its closed interior lie inside $|z|<R$. Parametrize it by $z=e^{it}$, $0\le t\le 2\pi$, so $dt=dz/(iz)$. For a positive integer $n$,The first contour integral is zero by the Cauchy integral theorem, since $f(z)z^{n-1}$ is holomorphic on the disk; the second is evaluated by the Cauchy derivative formula. ThereforeTaking $n=1$ gives the requested $\pi f^{\prime }(0)$.