= Solution
One form of the <Baire category theorem> says that a nonempty <complete metric space> cannot be a countable union of closed <nowhere dense sets>.
For each nonnegative integer $n$, put
$$
Z_n=\{z\in\mathbb C:f^{(n)}(z)=0\}.
$$
Because $f$ is an <entire function>, every <derivative> $f^{(n)}$ is continuous and $Z_n$ is closed. No $f^{(n)}$ is identically zero: if $f^{(n)}\equiv0$ for some $n$, repeated integration shows that $f$ is a <polynomial>. By the <identity theorem>, the zero set of the nonzero <holomorphic function> $f^{(n)}$ has empty interior. Thus each $Z_n$ is <nowhere dense set>[nowhere dense].
The <complex plane> is complete, so the <Baire category theorem> gives
$$
\mathbb C\ne\bigcup_{n=0}^{\infty}Z_n.
$$
Choose $z_0$ outside this union. Then $f^{(n)}(z_0)\ne0$ for every $n$, and hence every coefficient $f^{(n)}(z_0)/n!$ in the <Taylor series> about $z_0$ is nonzero. \b[Such a point $z_0$ therefore exists.]
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