= Solution
For an <entire function>, the <Cauchy derivative formula> on a positively oriented circle of radius $R$ about $z_0$ is
$$
\boxed{f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_{|\zeta-z_0|=R}
\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta,\qquad n=0,1,\ldots.}
$$
The resulting <Cauchy estimate> is $|f^{(n)}(z_0)|\leq n!M_R/R^n$, where $M_R$ is the maximum of $|f|$ on that circle.
The <Liouville theorem> says that a bounded <entire function> is constant. Indeed, if $|f|\leq M$ on the plane, the <Cauchy estimate> with $n=1$ gives $|f'(z_0)|\leq M/R$. Sending $R$ to infinity gives $f'(z_0)=0$ at every point, hence $f$ is constant.
For the growth assumption, apply the <Cauchy estimate> on circles centred at zero. For every integer $n>k$,
$$
|f^{(n)}(0)|\leq n!R^{k-n}\longrightarrow0\quad(R\longrightarrow\infty).
$$
All sufficiently high <Taylor series> vanish. Since the <Taylor series> of an <entire function> converges everywhere, \b[$f$ is a polynomial], of degree at most $\lfloor k\rfloor$ when $k\geq0$. If a negative finite exponent is interpreted for nonzero $z$, the same estimate forces every coefficient to vanish, giving $f=0$.
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