I am trying to understand the following equation: $$\sum_{k=0}^{\infty}\frac{a^k}{k!}\sum_{m=0}^{\infty}\frac{b^m}{m!} = \sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}a^kb^{n-k} = \sum_{n=0}^{\infty}\frac{(a+b)^n}{n!}$$
Its from the real & complex analysis book of Walter Rudin. It is a proof of the formula, $e^ae^b = e^{a+b}$
But I don't know how we get the second and third expressions. How are the limits changed?
The first equality follows from the multiplication theorem for the product of absolutely convergent series. (see Rudin's Principles theorem 3.50; actually only one of the series needs to converge absolutely). This justifies the first equality, (in the expression below the second equality) and we get
$$Exp(a)Exp(b) = \sum_{k=0}^{\infty}\frac{a^k}{k!}\sum_{m=0}^{\infty}\frac{b^m}{m!} = \sum_{k=0}^{\infty}\sum_{p=0}^{k}\frac{a^pb^{k-p}}{p!(k-p)!} =\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{p=0}^{k}\binom{k}{p}a^pb^{k-p} = \sum_{k=0}^{\infty}\frac{(a+b)^k}{k!} = Exp(a+b)$$ Besides the aforementioned theorem, only elementary manipulations and the binomial theorem were used.