Problem understanding the proof of Casorati-Weierstrass- Conway

Question

I am having trouble understanding why this $m+1$ is coming into the picture. Since $\frac{f(z)-c}{z-a}$ has a pole of order $m$, we have $$ \frac{f(z)-c}{z-a}=\frac{g(z)}{(z-a)^m}$$ for some analytic function $g$. Which implies $$\lim_{z\to a}(z-a)^{m+1}\frac{f(z)-c}{z-a}=\lim_{z\to a}(z-a)^{m}({f(z)-c})=\lim_{z\to a} (z-a)g(z)=0.$$ I am unable to understand why $\lim_{z\to a}(z-a)^{m+1}({f(z)-c})=0$ is written .