How to prove $x^ax^b = x^{a+b}$

It suffices to prove it for $x=e$ because then $x^ax^b=e^{a\ln x}e^{b\ln x}=e^{(a+b)\ln x}=x^{a+b}$.

$$e^ae^b=\sum_{n=0}^\infty\frac{a^n}{n!}\sum_{n=0}^\infty\frac{b^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{a^k}{k!}\frac{b^{n-k}}{(n-k)!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac1{n!}\binom nk a^kb^{n-k}=\sum_{n=0}^\infty\frac{(a+b)^n}{n!}=e^{a+b}$$ Here we use Cauchy's product formula, the series converge absolutely.