Here is an attempt.

We can write,

f(z) = (z-z_0)^m * g(z) for all z in D.

Then,

f'(z)=m(z-z_0)^{m-1} *g(z) + (z-z_0)^m*g'(z)

f'(z)=(z-z_0)^{m-1}*[mg(z)+(z-z_0)*g'(z)]

We need to show that the second function part does not attain a zero at z_0. Indeed! Substitute and see:

mg(z_0)+0*g'(z_0)=mg(z_0)!=0

Because m!=0 and g(z_0)!=0.

Thus, all the zero's belong to the first factor (z-z_0)^{m-1}.

Which is of order m-1.