Divergence of $z_{n+1}=(z_n)^2+c$ if there exists $n_0$ such that $z_{n_0} \geq 2$

Question

Let $z_{n+1}=(z_n)^2+c$ and suppose there exists $n_0$ such that $z_{n_0} > 2$. I want to show by induction that $$ \left|z_n\right| > 2^{n-n_0+1} \text{ for }n \geq n_0 $$ For $n=n_0$, we have $\left|z_{n_0}\right| > 2$. Suppose that it is true to a certain $N>n_0$ then $$ \left|z_{n+1}\right|=\left|\left(z_n\right)^2+c\right| > \left|\left|z_n\right|^2-\left|c\right|\right| > \left|2^{2n-2n_0+2}-\left|c\right|\right|=2^{n+1-n_0+1}\left|2^{n-n_0}-\frac{\left|c\right|}{2^{n-n_0+1}}\right| $$ How can I conclude from there since $c \in \mathbb{C}$ is unknown ?