Solution
= Solution
A <pole> of order $m\geq1$ at $a$ has finitely many negative powers in its <Laurent series>, the lowest being exactly $-m$:
$$
c_{-m}\neq0,\qquad c_j=0\quad(j<-m).
$$
Equivalently,
$$
\boxed{f(z)=(z-a)^{-m}h(z),\qquad h\text{ holomorphic near }a,\ h(a)\neq0.}
$$
Thus $(z-a)^m f(z)$ has a holomorphic extension with nonzero value at $a$, whereas multiplying by any smaller power does not remove the singularity.