On page 97 under the heading "Counting zeroes; the open mapping theorem" there is a second paragraph which goes like this:
In section 3 it was shown that if an analytic function $f$ had a zero at $z=a$ we could write $f(z)=(z-a)^mg(z)$ where $g$ is analytic and $g(a)\neq0$. Suppose $G$ is a region and let $f$ be analytic in $G$ with zeroes at $a_1,...,a_m$(where some of the $a_k$ may be repeated according to the multiplicty of the zero.) So we can write $f(z)=(z-a_1)(z-a_2)...(z-a_m)g(z)$ where $g$ is analytic on $G$ and $g(z)\neq0$ for any $z$ in $G$.
Here since $a_k$'s are repeated according to their multiplicity, then why not $f(z)=(z-a_1)^{n_1}(z-a_2)^{n_2}...(z-a_m)^{n_m}g(z)$. where $g$ is analytic with $g(z)\neq0$ for any $z$ in $G$ and $n_1,n_2,...n_m$ are respective multiplicities?
There's no effective difference. It's just less clutter in the expression.