Define $$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
Prove that the series converges for all $z\in\Bbb C$
Show that the series converges uniformly over any bounded subset of $\Bbb C$
The first part can easily be obtained by using the ratio test:$$\lim_{n\to\infty}\frac{1/n!}{1/(n+1)!}=\lim_{n\to\infty}\frac{(n+1)!}{n!}=\lim_{n\to\infty}n+1\to\infty$$
My question is about the second part, this is what I did:
We will set a sequence: $(u_n(z))_{n\in\Bbb N}$ where $u_n(z)$ is the $n^{th}$ term in the sum $u_n(z)=\frac{z^n}{n!}$.
Because $u_n(z)$ exists for all $z$ we can conclude that for the bounded set $V\subset \Bbb C$, $\sup\limits_{z\in V}(u_n(z))<\infty$ so I'll construct the sequence $(M_n)_{n\in \Bbb N}:M_n=|\sup\limits_{z\in V}(u_n(z))|$
From the first part I know that $\sum_nM_n$ converges, hence by Weierstrass M-test $e^z$ is uniformly converges.
I can't find flaws in the solution of the second part but it looks too easy and general, is my solution okay?
Also, is there a way to solve the second part without using Weierstrass M-test?
I really don't see how is it that you justify your claim “From the first part I know that $\sum_nM_n$ converges”. It is true though:$$M_n=\sup_{z\in V}\left|\frac{z^n}{n!}\right|=\frac{\left(\sup_{z\in V}|z|\right)^n}{n!}$$and therefore you can indeed use the first part.