Bounding a $C^k$ function on the unit disk In my reading there is the following "simple" claim:

If $u \in C^k(\overline{\mathbb{D}})$ and if all derivatives of $u$ up to order $k$ vanish on $\partial \mathbb{D}$, then for some $C>0$ it holds that $|u(z)| \leq C (1-|z|)^k$ on $\mathbb{D}$.

Any insights or references would be appreciated.
 A: This is more or less a proof.
Take $z\in \mathbb{D}$. Write it as $z=(x,y) = (rcos(\theta),rsin(\theta))$ with $0\leq r<1$.
Now define $g_\theta:\mathbb{R} \rightarrow \mathbb{C}$  by $g_\theta(\rho) = u(\rho cos(\theta),\rho sin(\theta))$.
I guess that $g_\theta$ is in $C^k([0,1])$ and, using Taylor around $\rho = 1$, there exists $r_\theta$ in $(0,1)$ such that $$ g_\theta(r) = g_\theta(1) + g_\theta ^{(1)}(1) (r-1)+ \frac{g_\theta ^{(2)}(1)}{2!}(r-1)^2 + \dots + \frac{g_\theta ^{(k-1)}(1)}{(k-1)!}(r-1)^{(k-1)} + \frac{g_\theta ^{(k)}(r_\theta)}{k!}(r-1)^k$$
Using that $r = |z|$ and the hypothesis that the derivatives vanish at the boundary:
$$ u(z)=u(rcos(\theta),rsin(\theta))=g_\theta(r) = \frac{g_\theta ^{(k)}(r_\theta)}{k!}(r-1)^k = \frac{g_\theta ^{(k)}(r_\theta)}{k!}(|z|-1)^k$$
Taking absolute value yields:
$$ |u(z)| = \left| \frac{g_\theta ^{(k)}(r_\theta)}{k!}\right| (1-|z|)^k \leq \frac{M}{k!} (1-|z|)^k $$
where $M$ is a uniform bound in $\theta$ for the the $k$-th derivative of $g_\theta$ in $[0,1]$, that exists because of the continuity of the derivatives of $u$ up to order $k$ on the compact set $\overline{\mathbb{D}}$.
