= Solution
Take a circle of radius $R\geq S$ about <zero>. The <Cauchy estimate> gives, for each integer $k>n$,
$$
|f^{(k)}(0)|\leq k!CR^{n-k}\longrightarrow0\quad(R\to\infty).
$$
Therefore all these derivatives vanish. The <Taylor series> of an <entire function> converges to it everywhere, and hence
$$
\boxed{f(z)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}z^k.}
$$
As usual $n$ is a nonnegative integer; if a real exponent is allowed, the <degree> is at most its integer part. A negative growth exponent forces $f=0$.
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