If $f$ has an essential singularity at point $a$, then $(z-a)^nf (z),\ \forall n$ has also an essential singularity at $a$.

Question

I almost prove this statement. Firstly, I want to prove it without Laurent series. My solution: Suppose that $(z-a)^nf (z)$ hasn't essential singularity at $a,$ for some $n$. This means that $a$ is a pole of $f$ or $a$ is an unessential singularity of $f$. So, $\displaystyle\lim_{z\to a} (z-a)^n f(z)=w\in \mathbb{C}\cup \{\infty\}$. If $w=\infty$ then $\displaystyle\lim_{z\to a} f(z)=\displaystyle\lim_{z\to a} \frac{(z-a)^nf(z)}{(z-a)^n}=\infty$. Thus, $a$ is a pole of $f$ (impossible). If $w\in \mathbb{C}\setminus\{0\}$, then obviously (in the same way) $\displaystyle\lim_{z\to a}f(z)=\infty$ (impossible). The only case left now is $w=0$. What can I say about this? I believe somehow we have to show that $\displaystyle\lim_{z\to a} f(z)\in \mathbb{C}$ (which is again impossible).