I am asked to prove that the function $\exp\left(z^2\right)$ is entire. Now, I know that it can be proven via the Cauchy-Riemann equations, or even by stating that the given function is given by the composition of two entire functions ($z^2$ and $e^z$). However, I am wondering if the following way is also acceptable.
$$\begin{aligned}\lim_{h\to 0}\frac{e^{\left(z+h\right)^2}-e^{z^2}}{h}&=e^{z^2}\cdot\lim_{h\to 0}\frac{e^{h^2+2zh}-1}{h}\xrightarrow[]{\text{L'Hospital}}2ze^{z^2}\cdot\lim_{h\to 0}e^{h^2+2zh}=2ze^{z^2}.\end{aligned}$$ This solution seems much easier than using the C-R equations. Is this valid? Do I need to prove that this solution is continuous everywhere? Thank you.
Your calculation shows that $f(z):=\exp(z^2)$ is complex differentiable everywhere in the open set $\mathbb{C}$. Thus, $f$ is holomorphic on $\mathbb{C}$. Finally, entire functions are precisely the functions which are holomorphic on $\mathbb{C}$.
The only possible issue here is the use of l'Hôpital's rule. If you are comfortable with it, you're alright.