f is an entire function with $f^n=0$ for some $n \in \mathbb{N}$. I need to show that f is polynomial of degree at most n-1.
I know that this question was asked before. But I need another approach to solve it. I don't want to use the taylor expansion or the gauchy integral. I need the most basic approach (using only the differentiability of f).
On
It can be proved by induction. If $f'=0$, then $f$ is constant. Now, suppose that $f^{(n)}=0\implies f$ is a polynomial whose degree is smaller than $n$. If $f^{(n+1)}=0$, then $(f')^{(n)}=0$. So,$$f'(z)=a_0+a_1z+\cdots+a_nz^n.$$Therefore, for some constant $C$,$$f(z)=C+a_0z+\frac12a_1z^2+\cdots+\frac1{n+1}a_nz^{n+1}.$$
On
According to Wikipedia, $f:\mathbb C \rightarrow \mathbb C$ entire means that there exists a sequence $(a_k)_{k \in \mathbb N} \subseteq \mathbb C$ such that $f$ can be written as a convergent series $$ f(z) = \sum_{k = 0}^\infty a_k z^k \text{ at every }z \in \mathbb C. \tag{1} $$
The derivatives of $f$ are also entire, and are given by $$ f^{(n)}(z) = \sum_{k = n}^\infty k(k-1)\ldots(k-n+1)a_k z^{k-n},\;\forall z \in \mathbb C. $$ Thus, $f^{(n)}=0$ only iff $a_k = 0$ for all $k \ge n$. By (1), this is equivalent to saying that $f$ is a polynomial function of degree at most $n-1$.
Just integrate repeatedly. $f^{(n-1)}(z)$ is a constant $c_1$; $f^{(n-2)}(z)=c_2+c_1z$ and so on.