Prove that $|e^{z^2}|\le e^{|z|^2}$. I need help proving that $|e^{z^2}|\le e^{|z|^2}$, where $z\in\mathbb{C}$.

Hint : $|e^z| = |e^x|$ , for any complex number $z=x+iy$.
Here is more details : 
\begin{equation} |e^{z^2}| = |e^{x^2 - y^2}| = e^{x^2 - y^2} \leq e^{x^2 + y^2} = e^{|z|^2} \end{equation}
