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

More generally, for any $z=x+iy\in\mathbb{C}$, we have that
$$|e^{z}| = |e^{x+iy}| = |e^{x}|\cdot |e^{iy}|=e^x\cdot 1\leq e^{|x|}\leq e^{|z|}.$$
