Schwarz Lemma of Complex Analysis 
Let be $f:B(0,1)\to B(0,1)$ holomorphic function such that
  $$f(0)=f'(0)=\cdots=f^{(n-1)}(0)=0$$ but $f^{(n)}(0)\neq 0.$ Show that
  $|f(z)|\le |z|^n,$ for every $z\in B(0,1)$ and $|f^{(n)}(0)|\le n!.$

I think that I will use the Schwarz Lemma, cause by the Lemma, I have that $|f(z)|\le |z|$, but I don't know how I can to use the conditions about the derivatives.
Any suggestion? Thanks.
 A: By the assumption that $f(0) = f'(0) = \dotsc = f^{(n-1)}(0) = 0$, the Taylor series of $f$ centred at $0$ is
$$f(z) = \sum_{k = n}^\infty a_k z^k.$$
Hence the function
$$g(z) = \frac{f(z)}{z^n} = \sum_{k = n}^\infty a_k z^{k-n} = \sum_{k = 0}^\infty a_{n+m}\cdot z^m$$
is holomorphic on $B(0,1)$.
By the maximum modulus principle, for $0 < r < 1$, we have
$$\lvert g(z)\rvert \leqslant \max \{ \lvert g(\zeta) : \lvert \zeta\rvert = r\} = \max \biggl\{ \biggl\lvert \frac{f(\zeta)}{\zeta^n}\biggr\rvert : \lvert \zeta\rvert = r\biggr\} \leqslant \frac{1}{r^n},\tag{1}$$
for $\lvert z\rvert \leqslant r$, since $\lvert f(\zeta)\rvert < 1$. We thus have
$$\lvert g(z)\rvert \leqslant \inf \{ r^{-n} : \lvert z\rvert \leqslant r < 1\} = 1$$
for all $z\in B(0,1)$.
Hence, we have
$$\lvert f(z)\rvert = \lvert z^n\cdot g(z)\rvert = \lvert g(z)\rvert\cdot \lvert z\rvert^n \leqslant \lvert z\rvert^n$$
on $B(0,1)$.
The estimate for $f^{(n)}(0)$ follows similarly from the integral formula:
$$f^{(k)}(z) = \frac{k!}{2\pi i} \int_{\lvert \zeta\rvert = r} \frac{f(\zeta)}{(\zeta-z)^{k+1}}\,d\zeta$$
for $\lvert z\rvert < r < 1$, whence
$$\lvert f^{(k)}(0)\rvert \leqslant \frac{k!}{2\pi} \cdot (2\pi r)\cdot \frac{1}{r^{k+1}} = \frac{k!}{r^k}.$$
Since that inequality holds for all $r < 1$, it follows that
$$\lvert f^{(k)}(0)\rvert \leqslant k!$$
for all $k\in \mathbb{N}$. Note that for this estimate, it was immaterial that $f$ has a zero at $0$, only that $\lvert f\rvert$ is bounded by $1$ is required.
