Schwarz Lemma – like exercise

There’s this exercise: let $\,f\,$ be analytic on $$D:=\{z\;\;;\;\;|z|<1\}\,\,,\,|f(z)|\leq 1\,\,,\,\,\forall\,z\in D$$ and $\,z=0\,$ a zero of order $\,m\,$ of $\,f\,$.

Prove that $$\forall z\in D\,\,,\,\,|f(z)|\leq |z|^m$$

My solution: Induction on $\,m\,$: for $\,m=1\,$ this is exactly the lemma of Schwarz, thus we can assume truth for $\,k<m\,$ and prove for $\,k=m>1\,$ . Since $\,f(z)=z^mh(z)\,\,,\,h(0)\neq 0\,$ analytic in $\,D\,$ , put
$$g(z):=\frac{f(z)}{z}=z^{m-1}h(z)$$

Applying the inductive hypothesis and using Schwarz lemma $\,\,(***)\,\,$ we get that
$$|g(z)|=\left|\frac{f(z)}{z}\right|=|z|^{m-1}|h(z)|\stackrel{ind. hyp.}\leq |z|^{m-1}\Longrightarrow |f(z)|\leq |z^m|$$
and we’re done…almost: we still have to prove $\,|g(z)|\leq 1\,$ for all $\,z\in D$ in order to be able to use the inductive hypothesis and this is precisely the part where I have some doubts: this can be proved as follows (all the time we work with $\,z\in D\,$):

$(1)\,\,$ For $\,f(z)=z^mh(z)\,$ we apply directly Schwarz lemma and get
$$|f(z)|=|z|^m|h(z)|\leq |z|\Longrightarrow |z|^{m-1}h(z)|\leq 1$$
And since now the function $\,f_1(z)=z^{m-1}h(z)\,$ fulfills the conditions of S.L. we get

$(2)\,\,$ Applying again the lemma,
$$|f_1(z)|=|z|^{m-1}|h(z)|\leq |z|\Longrightarrow |z^{m-2}h(z)|\leq 1$$and now the function $\,f_2(z):=z^{m-2}h(z)\,$ fulfills the conditions of them lemma so…etc.

In the step$\,m-1\,$ we get
$$|z||h(z)|\leq |z|\Longrightarrow {\color{red}{\mathbf{|h(z)|\leq 1}}}\,$$
and this is what allows us to use the inductive hypothesis in $\,\,(***)\,\,$ above.

My question: Is there any way I can’t see right now to deduce directly, or in a shorter way, that $\,|h(z)\leq 1\,$ ?

Solutions Collecting From Web of "Schwarz Lemma – like exercise"