If $f$ is holomorphic on $\mathbb D$, $|f(z)| < 1;\forall z$, $f$ has a zero of order 2 at $0$, then $|f(z)|\leq |z|^2;\forall z$ So my first step is to cite Schwarz's Lemma, so we have that $|f(z)|\leq |z|$, which is close to what we want. But I'm not sure how to bridge the gap to $|z|^2$.
There is a theorem which says that, since $f'(0)=0$, there are $\delta ,\epsilon$ such that each point in $N_\epsilon (0)\setminus \{0\}$ has exactly $2$ preimage points (which are also simple points of $f$) in $N_\delta (0)\setminus \{0\}$. I wonder if this could be helpful.
One way to prove this would be to prove that $\displaystyle g(z)=\frac{f(z)}{z}$ satisfies Schwarz's Lemma too. We already have that $g(0)=0$, but I don't see a way to prove that $|g|\leq 1$ on $\mathbb{D}$. I'm not even sure if that is a true statement.
I have seen a similar statement in another question on this website, and they suggested using the Maximum Modulus Theorem to prove this, but I don't see how exactly that applies. $f$ has a maximum on the unit circle. $g$ would also need to take a maximum there, since $g(0)=0$. So what?
We know that $\displaystyle |g(z)|\leq \frac{1}{\inf \limits _{z\in \mathbb{D}}|z|}$, but this says nothing really.
 A: 
Let $D$ be the open unit disc and suppose $f:D\to\Bbb{C}$ is a holomorphic function with a zero of order $n$ ($n\geq 1$ an integer) at the origin. If for all $z\in D$, we have $|f(z)|\leq 1$, then

*

*For all $z\in D$, we have $|f(z)|\leq |z|^n$.

*If there is a $z_0\in D\setminus\{0\}$ such that $|f(z_0)|=|z_0|^n$, then there is a $\lambda\in S^1$ such that for all $z\in D$, we have $f(z)=\lambda z^n$.



Proof 1.
Consider the function $g(z)=\frac{f(z)}{z^n}$; strictly speaking this is a-priori only defined on $D\setminus\{0\}$ and holomorphic there, but because $f$ has a zero of order $n$, it follows $g$ has a holomorphic extension to $D$. Now, on the circle $\{z\,:\, |z|=r\}$ (where $0<r<1$) we have
\begin{align}
|g(z)|&=\frac{|f(z)|}{|z|^n}\leq\frac{1}{r}.
\end{align}
Hence, by the maximum modulus principle, it follows that for all $|z|\leq r$, we again have the same inequality. Now, fix a point $z\in D$, and consider any $r$ such that $|z|<r<1$. Then, we have $|f(z)|\leq \frac{|z|^n}{r}$ by the above inequality; so by taking the limit $r\to 1^-$, we get $|f(z)|\leq |z|^n$. Since $z\in D$ was arbitrary, this completes the proof of the first statement.
For the second statement, fix $|z_0|<r<1$. Then, $g$ is bounded by $1$ on the closed disc $\overline{D}_r$, and it attains the maximum value of $1$ at the interior point $z_0$. Hence, by the maximum modulus principle, $g$ must be constant on $\overline{D}_r$; and this constant (which equals $g(0)$) has absolute value $1$ because $|g(0)|=|g(z_0)|=\frac{|f(z_0)|}{|z_0|^n}=1$. So, $f(z)=g(0)\cdot z^n$ for all $|z|\leq r$. Finally, since $r$ was arbitrary, the same equality holds for all $z\in D$.

Proof 2.
Given $f$ as in the theorem, define $g(z)=\frac{\phi(z)}{z^{n-1}}$; this extends holomorphically to $D$ with $g(0)=0$. Fix $0<r<1$. Then, on the circle $|z|=r$, we have
\begin{align}
|g(z)|&=\frac{|f(z)|}{|z^{n-1}|}\leq\frac{1}{r^{n-1}}.
\end{align}
Therefore, by the maximum modulus principle, the same inequality holds for all $|z|\leq r$. Since $r$ is arbitrary, it follows that for all $z\in D$, we have $g(z)\leq 1$, and $g(0)=0$. So, $g$ satisfies the hypotheses of the usual Schwarz lemma, hence, $g(z)\leq |z|$ in $D$, which implies $|f(z)|\leq |z|^n$ in $D$.
For the second statement, we have $|g(z_0)|=|z_0|$ for some $z_0\in D\setminus\{0\}$, so there is a $\lambda\in S^1$ such that $g(z)=\lambda z$, and hence $f(z)=\lambda z^n$ for all $z\in D$.

Remarks about the Proofs.
Note that the method of the first proof is exactly how the usual Schwarz lemma ($n=1$) is established. The second proof, I wouldn't really consider as a distinct proof; it's pretty much induction and the maximum modulus principle (the key to the first proof) mashed together into one approach.
