Suppose $f$ is holomorphic on an open disc $D$ centered at $a$ of radius $R$. Suppose that $|f(z)|\leq M$ on $D$. Suppose further that $f^{(n)}(a)=0$ for $0\leq n<k$. Then show that $|f(z)|\leq M\left(\frac{r}{R}\right)^k$ for all $|z-a|\leq r<R$.
My progress is as follows. By writing $f$ as a power series $\sum c_i (z-a)^i$, we know that the first $k$ coefficients are zero. Then, by using triangle inequality and Cauchy's Inequalities/Estimates, we can further bound the coefficients $c_i$. In turn, the best that I am able to show is that $|f(z)|\leq M\sum_{n=k}^{\infty}\left(\frac{r}{R-\epsilon}\right)^k$ for $|z-a|\leq r<R-\epsilon<R$. Obviously, this is not nearly strong enough, but frankly at each of my steps I don't seem to be losing any sharpness at all. So I am at a loss as to how to proceed. Any and all help would be appreciated.
Remark: This question was posted by @lagicol as part of a question that contained 2 questions. I suggested that he split the question into 2 questions, but somehow he misunderstood and deleted the question, after I have answered the first part of the question. So my efforts were in vain. I am posting my answer here as it might be useful to others.
@lagicol: If you are reading this, please restore your original question, and only move the second part of that question into a new post, without deleting the original post.
Without loss of generality, set $a=0$. As you have observed we have $$ f(z)=c_kz^k+c_{k+1}z^{k+1}+\ldots, $$ and so $$ g(z)=\frac{f(z)}{z^k}=c_k+c_{k+1}z+\ldots, $$ is an analytic function in the disk $D$. Consider the close disk $D_r$ of radius $r>0$, centred at the origin. Since $g$ must take its maximum (in modulus) at the boundary, we have $$ |g(z)|\leq|g(w)|, $$ for all $z\in D_r$ and for some $w$ with $|w|=r$. That is, $$ |g(z)|\leq|g(w)|=\frac{f(w)}{|w|^k}\leq\frac{M}{r^k}, $$ and sending $r\to R$ yields $$ |g(z)|\leq\frac{M}{R^k}, $$ or $$ |f(z)|\leq M\frac{|z|^k}{R^k} , $$ which is the desired result.