Show that if a holomorphic function $f : U → \Bbb C$ has a pole of order $N$ at $0$, then $|f (z)| |z|^{N−1} \to +\infty$ when $z \to_\neq 0$.
I did the following :
We know that since $f$ is analytic, by the isolated zeros principle, we can find a radius $r$ such that $f(z)\neq0$ $\forall z \in D(0,r)$, we will have that $$0<\alpha<\lvert f(z) \rvert \lvert z \rvert^N \quad \forall z \neq 0 \in D(0,r) $$
Then $$0<\frac\alpha{\lvert z\rvert}<\lvert f(z) \rvert \lvert z \rvert^{N-1} \quad \forall z \neq 0 \in D(0,r)$$
When $z$ goes to $0$, $\rvert z \rvert$ does the same and $\frac\alpha{\lvert z\rvert}$ goes to $+ \infty$, so $\lvert f(z) \rvert \lvert z \rvert^{N-1}$ that is bigger does the same (comparison criterion).
The problem is that the same proof can be done to show that $\lvert f(z) \rvert \lvert z \rvert^N \to + \infty$ which is false because $\lvert f(z) \rvert \lvert z \rvert^N$ is bounded near $0$. Can someone see this is wrong ?