Upper bound on complex integral If $f(z)=\sum_{n=0}^{\infty}c_nz^n$ and we know $$c_k=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z^{k+1}}dz$$ for $\gamma$ a circle of radius r centred at the origin, traversed once in the positive direction, how can it be shown that:
$$r^k|c_k|\leq \max\limits_{|z|=r}\{|f(z)|\}$$
holds for $r<R$ the radius of convergence of $f(z)$?
I assume the formula to be used here is that 
$$\left|\int_\gamma f(z)dz\right|\leq length(\gamma)\max\limits_{z\in \gamma^*}\{|f(z)|\}.$$
Since $\gamma$ is a circle traversed once, the length must be $2\pi r$. This is where I am stuck. I have tried rearranging the expression for $c_k$ but nothing gives an expression inside the modulus with just the integral of $f(z)$. How else can I manipulate this expression? 
The furthest I have managed is $|c_k|=\left|\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z^{k+1}}dz\right|=\frac{1}{|2\pi i|} \left|\int_\gamma \frac{f(z)}{z^{k+1}}dz\right|$ which doesn't at all look like multiplying by $r^k$ will get the $f(z)$ alone in the integral.

By the ML-inequality,
$$|c_k| \le \frac{1}{2\pi} \max_{|z| = r}\cdot \left|\frac{f(z)}{z^{k+1}}\right|\cdot 2\pi r = r^{-k}\max_{|z| = r}|f(z)|,$$
and hence $r^k|c_k| \le \max_{|z| = r} |f(z)|$.
