Since $g(z)$ is a polynomial of degree $n$, so is its Taylor series about the point $z=c$. Note that $g^{(n)}(z) = n!$, so that the coefficient of $(z-c)^n$ in this series is $1$:
$ g(z) = g(c) + g'(c)(z-c) + \cdots + \frac{g^{(n-1)}(c)}{(n-1)!} (z-c)^{n-1} + (z-c)^n. $
Suppose that, for some $m \in \{0,1,2,\cdots,n-1\}$ and $\epsilon > 0$, we know that
$ \left|g^{(k)}(c)\right| \leq \epsilon $
for all $0 \leq k \leq m$. It then follows from the triangle inequality that the head of the polynomial satisfies
$ \begin{align} \left|g(c) + g'(c)(z-c) + \cdots + \frac{g^{(m)}(c)}{m!} (z-c)^{m}\right| &\leq \epsilon \sum_{k=0}^{m} \frac{|z-c|^k}{k!} \\ &< \epsilon e^{|z-c|}. \tag{1} \end{align} $
The rest of the polynomial
$ h(z) = \frac{g^{(m+1)}(c)}{(m+1)!} (z-c)^{m+1} + \cdots + \frac{g^{(n-1)}(c)}{(n-1)!} (z-c)^{n-1} + (z-c)^n $
has a zero of multiplicity at least $m+1$ at $z=c$, and, since the zeros of analytic functions are isolated, this is the only zero of $h(z)$ in the closed disk $|z-c| \leq \delta$ for all $\delta > 0$ small enough. The circle $|z-c| = \delta$ is compact and $h(z) \neq 0$ there, so we can also find a $\lambda > 0$ such that
$ |h(z)| \geq \lambda > 0 \qquad \text{on}\,\,\, |z-c| = \delta. \tag{2} $
Now, if we choose
$ \epsilon \leq \lambda e^{-\delta} $
then we can deduce from $(1)$ and $(2)$ that
$ \begin{align} \left|g(c) + g'(c)(z-c) + \cdots + \frac{g^{(m)}(c)}{m!} (z-c)^{m}\right| &< \epsilon e^{\delta} \\ &\leq \lambda \\ &\leq |h(z)| \end{align} $
on the circle $|z-c| = \delta$.
We may now apply Rouché's to conclude that for any fixed $\delta > 0$ we can find an $\epsilon > 0$ small enough so that the polynomial $g(z)$ has at least $m+1$ zeros in the disk $|z-c| < \delta$.