I was rereading the prologue to Real and Complex Analysis by Rudin (in image) and I realized I never really understood why we have
the first and second equality after$\sum_{k = 0}^{\infty}\frac{a^k}{k!}\sum_{m = 0}^{\infty}\frac{b^m}{m!} $ and where does the absolute convergence of (1) play a role.
So could somebody provide some extra steps and maybe I will understand?
This first and second formula are nothing but $$ (1+a+{a^2\over2}+...)(1+b+{b^2\over2}+...) =1\times1+a\times1+1\times b+a\times b+1\times{b^2\over2}+{a^2\over2}\times1\\ =1+(a+b)+{(a+b)^2\over2}+...\\ $$ The $$ \sum{1\over n!}\sum{n!\over(n-k)!k!} $$ uses $n!/n!=1$ to get an $n\choose k$ in the sum to show that the coefficients of $a^k b^{n-k}$ are the binomial expansion of $(a+b)^n$.
A proof that absolute convergence allows the summations to be exchanged can be found online here. The proof is quite long so I won't copy it all out here. It relies on another theorem which I wasn't able to locate at that web site.
The author also constructs a counterexample of a series which converges, but not absolutely, and shows that the order of summation affects the limit value.