If $f$ has a pole of order $m\ge 1$ in $a$ then $\lim_{z\to a}\vert f(z)\vert=\infty.$ Let $f$ be holomorphic on $\Omega\setminus{\\{a\\}}$ where $a\in \Omega$ and $\ \Omega$ is an open set. $f$ has a pole of order $m\ge 1$ in $a$ if in a deleted neighborhood of $a$ I can write $$f(z)=\sum_{k=1}^m \frac{c_{-m}}{(z-a)^k}+\sum_{k=0}^{\infty} c_k(z-a)^k$$ for $c_{-m}\ne 0.$ Now I would like to prove that it implies that $\lim_{z\to a}\vert f(z)\vert=\infty.$ Denote $R$ the radius of convergence of $\sum_{k=0}^{\infty} c_k(z-a)^k$ and $0<h<R$ so that $\vert z-a\vert\le h,$ Then we have $(z-a)^mf(z)=c_{-m}+c_{-m+1}(z-a)+\ldots+\sum_{k=0}^{\infty}c_k(z-a)^{k-m}.$ Not sure how can continue I would like something $\vert f(z)\vert\times\vert z-a\vert^m\ge A_m$ where $A_m$ would be positive constant. I tried the revers inequality, but I have something like $\vert c_{-m}-h(\vert c_{-m+1}\vert\ldots)\vert$

But you already have it...almost:

$$|f(z)|=\frac1{|z-a|^m}\left|a_{-m}+a_{m+1}(z-a)+\ldots\right|$$

and now just observe that

$$\left|a_{-m}+a_{m+1}(z-a)+\ldots\right|\xrightarrow[z\to a]{}|a_{-m}|$$