justification on changing indexes in double sum I was wondering what is the justification for this step(changing the indexes)$\displaystyle\sum_{n=0}^{\infty}\frac{a^{n}}{n!}\sum_{m=0}^{\infty}\frac{b^{m}}{m!}=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n=0}^{k}\frac{k!}{n!(k-n)!}a^{n}b^{k-n}$ , in Rudin's Real and complex analysis prolog to show $(\exp{a})( \exp{b})=\exp{(a+b)}$, is it the same principle that use fubini's theorem for integrals, I mean that one that says if given the domain of integration D=AxB=ExF then I can do something like $\int_{D}=\int_{A}\int_{B}=\int_{E}\int_{F}$ , I would appreciate any hint or reference to this, thanks in advance.
 A: If we let $f_n = \frac{a_n}{n!}$ and $g_m = \frac{b_m}{m!}$, you're asking why 
$$\displaystyle\sum_{n=0}^{\infty} f_n \sum_{m=0}^{\infty} g_m = \sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n=0}^{k} k! f_n g_{k-n} = \sum_{k=0}^{\infty}\sum_{n=0}^{k}f_n g_{k-n}$$
The reason is that the expression $$\sum_{n=0}^{\infty} f_n \sum_{m=0}^{\infty} g_m $$ is adding up the following numbers row-by-row,
$$
\begin{matrix}
f_0 g_0 & f_0 g_1 & f_0 g_2 & f_0 g_3 & \cdots \\
f_1 g_0 & f_1 g_1 & f_1 g_2 & f_1 g_3 & \cdots \\
f_2 g_0 & f_2 g_1 & f_2 g_2 & f_2 g_3 & \cdots \\
f_3 g_0 & f_3 g_1 & f_3 g_2 & f_3 g_3 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix},
$$
while the sum $$\sum_{k=0}^{\infty}\sum_{n=0}^{k}f_n g_{k-n}$$ is adding up the same numbers diagonal-by-diagonal.

Added: The answer above intentionally ignores questions of convergence.  It assumes that the OP's question must be about the algebra of why the equality is true, since if there were problems with convergence the equality wouldn't be asserted in Rudin.  However, for those concerned about convergence issues here, Merten's theorem can be applied, as wnoise points out below.
