Show that f is a polynomial of degree $\le n$ From Basic Complex Analysis 3rd ed. 1.5 #20
Let f be an analytic function on an open connected set A and suppose that $f^{n+1} (z)$ (the n+1st derivative) exists and is zero on A. Show that f is a polynomial of degree $\le n$.  
This statement being true seems very intuitive to me but I'm struggling with how I might show this. My question lies in whether or not 0 is a part of the natural numbers for this problem (couldn't find anything in the book). It seems that if it is the problem is saying f'(z)=0 or if 0 is not included f''(z)=0. 
My attempt at a solution (assuming $0 \in \mathbb{N}$)  
Let $f(z)=cz^m$ where c and m are any constant 
Then $f'(z)=mcz^{m-1}=0$ 
From here I note that $m=0 \le \mathbb{N}$ 
Is this sufficient? 
 A: Hint: Since $f^{(n+1)}$ is zero on $A$ so $f^{(n)}$ is constant on $A$. Now what can you say about $f^{(n-1)}$, and then $f^{(n-2)}$ and ...
A: Since $f$ is analytic on A, then for $a\in A $ locally we have 
$$f(z) = \sum_{j=0}^{\infty} a_j (z-a)^j,~~~~~\text{for }~~|z-a|<r.$$

We have, 
$$ f^{(m)}(z) = \sum_{j=0}^{\infty} a_j 
\frac{d^m}{dz^m}[(z-a)^j],~~~~~\text{for }~~|z-a|<r.$$
But $$\frac{d^m}{dz^m}[(z-a)^j] = 0 ~~~~\text{if}~~j<m$$
and $\text{if}~~j\ge m$ we have, $$\frac{d^m}{dz^m}[(z-a)^j] = j(j-1)\cdots (j- m+1) (z-a)^{j-m}=\frac{j!}{(j-m)!}(z-a)^{j-m}~~$$
Whence, by assumption, $f^{(n+1)}(z)=0 $ on A  we have get,
$$ f^{(n+1)}(z) = \sum_{j=n+1}^{\infty} a_j 
\frac{j!}{(j-n-1)!}(z-a)^{j-n-1} =0,~~~~~\text{for }~~|z-a|<r.$$
which is possible only if $$ a_j =0 ~~~\text{for }~~~j\ge n+1$$
Hence, 
$$f(z) = \sum_{j=0}^{\infty} a_j (z-a)^j= \sum_{j=0}^{n} a_j (z-a)^j~~~~~\text{for }~~|z-a|<r.$$
That is $f$ and $p(z)=\sum_{j=0}^{n} a_j (z-a)^j$ coincide on $D(a,r)\subset A$ and A is connected by analytic extension we get that, 
$$ 
$$f(z) = \sum_{j=0}^{n} a_j (z-a)^j~~~~~\text{ for }~~ \forall~~z\in A.$$
