Suppose that $f$ has a pole of order $m$ at $a$ and $g$ is holomorphic in $D(a;r)$ (open disc centered at a with radius $r>0$). Then at $a$ the function $fg$
(i) has pole of order $n-m$ if $g$ has zero of order $n$ ($n < m$) at $a$.
(ii) has removable singularity if $g$ has zero of order $n$ ($n > m$) at $a$.
Argument:
Since $f$ has a pole of order $m$, we can write it as $$ f(z) = \lim_{z \to a} \frac{D}{(z-a)^m}, $$ where $D$ is a non-zero constant.
Similarly, $$ g(z) = \lim_{z \to a} {(z-a)^n}h(z), $$ where $h(z) \neq 0$.
Hence, \begin{align*} fg(z) &= \lim_{z \to a} \frac{D}{(z-a)^m} \times \lim_{z \to a} (z-a)^n h(z) \\ &= \lim_{z \to a} \frac{D}{(z-a)^m}{(z-a)^n}h(z). \end{align*}
For (i), since $m > n$, we have $(z-a)^{m-n}$ in the denominator, hence $fg$ has a pole of order $m-n$.
For (ii), since $m < n$, we have $(z-a)^{n-m}$ in the numerator, hence $fg$ should have zero of order $n-m$.
But why is there a removable singularity? Please explain.
In case (ii), the product $fg$ isn't defined at $z=a$, since $f$ isn't.
Hence, at least formally, $fg$ has a singularity at $z=a$, but your computation shows that it is removable. (Putting $fg(a) = 0$ gives you a holomorphic function with a zero of order $n-m$ at $z=a$.)