Upper bound for $|f(0)|$ based on its behavior within the unit disk. Function $f(z)$ is holomorphic at least within the unit disk. 
If for all arguments $|z|<1$ we have $|f(z)| \le 1$, and there exists a zero of order $n$ at $z=z_0$, then show that $|f(0)| \le |z_0|^n$.
How should I tackle this problem? Can it be generalized if for $|z|<R$ we have $|f(z)| \le M$ ?

So far I only understand that there exists a holomorphic $g(z)$ with $g(z_0) \ne 0$ such that $f(z) = (z-z_0)^n g(z)$,then $|f(z)| = |z-z_0|^n |g(z)| \le 1$ and moreover $|f(0)| = |z_0|^n |g(0)|$, but that doesn't seem useful since apparently I know too little about $g(z)$. In particular why should $|g(0)| \le 1$?
 A: The usual method to prove this kind of estimates is the Schwarz lemma and its variants. The Schwarz lemma says that a holomorphic function $g\colon \mathbb{D} \to \mathbb{D}$ with $g(0) = 0$ satisfies $\lvert g(z)\rvert \leqslant \lvert z\rvert$ for all $z\in\mathbb{D} = \{z\in \mathbb{C} : \lvert z\rvert < 1\}$, and if equality holds at any $z\neq 0$, then $g$ is a rotation, $g(z) = e^{i\varphi}z$ for some $\varphi \in [0,2\pi)$. It generalises to holomorphic functions $g\colon \mathbb{D}\to \mathbb{D}$ having a zero of order $m$ at $0$: We consider the function $$k(z) = \frac{g(z)}{z^m},$$ which by assumption has a removable singularity at $0$, which we presume removed. For $0 < r < 1$, the premise $\lvert g(z)\rvert \leqslant 1$ yields $\lvert k(z)\rvert \leqslant \frac{1}{r^m}$ on the circle $\lvert z\rvert = r$, and by the maximum modulus principle that estimate holds for $\lvert z\rvert \leqslant r$. Letting $r\to 1$, we obtain $\lvert k(z)\rvert \leqslant 1$ for all $z\in\mathbb{D}$, and hence $\lvert g(z)\rvert \leqslant \lvert z\rvert^m$ for $\lvert z\rvert < 1$. If $\lvert k(z)\rvert = 1$ for any $z\in\mathbb{D}$, that is if $\lvert g(z)\rvert = \lvert z\rvert^m$ holds for some $z\neq 0$ or $\lvert g^{(m)}(0)\rvert = m!$, then it follows that $k$ is constant with modulus $1$, $k(z)\equiv e^{i\varphi}$, and hence we have $g(z) = e^{i\varphi}\cdot z^m$ in that case.
To apply the above, compose $f$ with an automorphism of the unit disk to move the zero to $0$.
So we consider the function
$$g(z) = f\left(\frac{z+z_0}{1 + \overline{z_0}\cdot z}\right).$$
$g$ is holomorphic (at least) on the open unit disk, we have $\lvert g(z)\rvert \leqslant 1$ for all $z$ with $\lvert z\rvert < 1$, and $g$ has a zero of order $n$ at $0$. By the above, $\lvert g(z)\rvert \leqslant \lvert z\rvert^n$. Now $f(0) = g(-z_0)$, whence we obtain the desired $$\lvert f(0)\rvert \leqslant \lvert -z_0\rvert^n = \lvert z_0\rvert^n.$$
The result can be generalised to $\lvert f(z)\rvert \leqslant M$ for $\lvert z\rvert < R$ by considering the function $h(z) = \frac{1}{M} f(R\cdot z)$. If $f$ has a zero of order $n$ at $z_0$, then $h$ has a zero of order $n$ at $\frac{z_0}{R}$, and the above yields $\lvert h(0)\rvert \leqslant \left\lvert \frac{z_0}{R}\right\rvert^n$, so $$\lvert f(0)\rvert = \lvert M\cdot h(0)\rvert \leqslant M\cdot R^{-n}\cdot \lvert z_0\rvert^n.$$
