Proof that if polynomial in $\mathbb{Z}[x]$ has a rational root $a/b$ and $k$ is an integer, then $a-kb|f(k)$.

Question

I'm currently learning about rings and polynomials and I was asked to prove that thing in the title.

So far I can prove that $f(x)$ can be written as $\sum_{i=0}^{n} z_i (x-k)^i$ with $z_0 ,...,z_n \in \mathbb{Z}$ by applying the division algorithm. Consequently $0=f(a/b)=\sum_{i=0}^{n} z_i (\frac{a}{b}-k)^i = \sum_{i=0}^{n} z_i \frac{(a-kb)^i}{b^i} = z_0 +(\sum_{i=1}^{n} z_i \frac{(a-kb)^{i-1}}{b^i})(a-kb)$ being $f(k)=z_0$. What I'm unable to prove is $\sum_{i=1}^{n} z_i \frac{(a-kb)^{i-1}}{b^i} \in \mathbb{Z}$.

Thanks in advance.