Prove that $\exists k\geq 0$ such that $f^n(0)=0$ for all $n\geq k$ Let $f $be an entire function.Suppose that for each $a\in \Bbb R ,\exists c_n$ in the expansion $ f(z)=\sum c_n(z-a)^n$ for some $n$ such that $c_n=0$.
Then prove that 

$\exists k\geq 0$ such that $f^n(0)=0$ for all $n\geq k$.

Now $c_n=\dfrac{f^n(a)}{n!}$. But it is not helping .Any hint will be highly beneficial.
 A: For each $a \in \mathbb{R}$ there exists some $n \in \mathbb{N}$ such that $c_n = \frac{f^n (a)}{n!} = 0$ which implies that $f^n (a) = 0$. 
Now comes the key to this exercise : there are countably many derivative functions, yet we have an uncountable amount of values for which the derivative functions vanish. This implies that there must exist some $n \in \mathbb{N}$ such that $f^n (x) = 0$ for $x \in A$ such that $A \subset \mathbb{R}$ is uncountable.
Since $f^n (x) = 0$ for an uncountable amount of distinct points, and an uncountable set of points always has a limit point, then it follows by the identity theorem that $f^n (x) = 0$ for all $x \in \mathbb{C}$. 
Since $f^n (x) = 0$ for all $x \in \mathbb{C}$ it follows that $f^n (x)$ is a constant function and hence its remaining derivatives all vanish.
When I first saw this exercise quite some time ago, I was rather overwhelmed. I suggest you read through my answer and think deeply about why certain steps are true. In particular I suggest thinking through the uncountability argument used, and pay close attention to the use of the identity theorem  and the way in which its conditions are met.
