Let $0<r<1$, $f:U_r\setminus\{0\}\to\mathbb{C}$ holomorphic. Let $\alpha <1,\; M\ge 0$ such that $$|f(z)|\le\frac{M}{|z|^{\alpha}}$$for all $z\in U_r(0)\setminus \{0\}$. Prove that $0$ is a removable singularity of $f$.
I tried to prove it but I have no idea how to do it exactly. I would start with the Laurent-series of $f$ in $z_0=0$ : $$f(z)=\sum_{k=-\infty}^{\infty}a_kz^k$$ and the goal should be the prove $f(z)=\sum_{k=0}^{\infty}a_kz^k$, i.e. $a_k=0$ for all $k<0$. First I have that $$\left|\sum_{k=-\infty}^{\infty}a_kz^k\right|\le\frac{M}{|z|^{\alpha}}$$for all $z\in U_r(0)\setminus \{0\}$, but I don't know how to continue.
Therefore my question is: How to prove it (other ideas are welcome too) or how to continue?
Edit: With zhw's great hint I get (I wrote that as a comment, too) : I have: $|zf(z)|\le \frac{|z|M}{|z|^\alpha} =M|z|^{1-\alpha}$ with $1-\alpha>0$. It follows $zf(z)\to 0$ for $z\to 0$. Therefore either $f$ has a pole in $0$ of order 1 or $0$ is a removable singularity of $f$. Is it correct? But then I am stuck again. If I assume, $0$ is a pole, I don't know where I get a contradiction.
Hint: Consider $\,zf(z).\,\,\,$