Exponential Power Series Expansion

[Lucid Dreamer]

I want to show that e^x e^x = e^{2x} using a power series expansion. So I start with

\sum_{n=0}^\infty \frac{x^n}{n!} \sum_{m=0}^\infty \frac{x^m}{m!}
\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^n}{n!} \frac{x^m}{m!}
\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^{m+n}}{m!n!}

I am at a loss of where to go from here. I want to reduce the last expression down to \sum_{n=0}^\infty \frac{(2x)^n}{n!} but I am not sure of how to get rid of one of the summations.

 

[Lucid Dreamer]

I think I got it...in case anyone was interested.

I can use a substitution on the indicies of the double series as follows. Let m=q and n=p-q, with the condition that q \le p. This gives

\sum_{p=0}^\infty \sum_{q=0}^p \frac{x^q}{q!}\frac{x^{p-q}}{(p-q)!}
Which one recognizes as a binomial expansion. Where \frac{(x+x)^n}{n!} = \sum_{k=0}^n \frac{1}{k!(n-k)!} x^k x^{n-k}. Thus our double series reduces down to

\sum_{p=0}^\infty \frac{(x+x)^p}{p!} = e^{2x}