e^x e^y = e^(x+y) from power series

Cauchy product. Both series converge absolutely, so their product is the series with general term $c_n=\sum_{k=0}^na_kb_{n-k}$, where $a_k=\frac{x^k}{k!}$ and $b_j=\frac{y^j}{j!}$.

Computation.
$$c_n=\sum_{k=0}^n\frac{x^k}{k!}\cdot\frac{y^{n-k}}{(n-k)!}=\frac1{n!}\sum_{k=0}^n\binom nkx^ky^{n-k}=\frac{(x+y)^n}{n!}$$
by the binomial theorem.

Conclusion. $\exp(x)\exp(y)=\sum_n\frac{(x+y)^n}{n!}=\exp(x+y)$.

Consequences. $\exp(x)\exp(-x)=1$, so $\exp$ never vanishes, and $\exp(1)=e$ gives $\exp(n)=e^n$: this is the rigorous way to show that the series is the exponential function.