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D={z : |z| $\lt$ 1}. $\mathit f$ : D$\to$ $\bar D$ is analytic and have a zero of order n at z=0. Prove that |f(z)| $\le$ $|z|^n$ $\forall$ z $\in$ D. Further show that |$f^n$(0)| $\le$ n!.

Since f is analytic and have a zero of order n at z=0 it can be expressed as f(z)= $z^n$g(z) where g is analytic and g(0) $\neq$ 0. Somehow I have to show that |g(z)| $\le$ 1 on D. But I am not being able to understand how to proceed further. Any kind of suggestion is highly appreciated.

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