Prove $\exp(x+y) = \exp(x) \exp(y)$ for $\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$

\begin{align} \exp(x+y)&=\sum_n\frac{(x+y)^n}{n!} \\\\ &=\sum_{n}\frac{1}{n!}\sum_{a+b=n} {n \choose a} x^ay^b \\\\ &= \sum_{n}\frac{n!}{n!}\sum_{a+b=n}\frac{x^a}{a!}\frac{y^b}{b!} \\\\ &= \sum_{a,b} \frac{x^a}{a!}\frac{y^b}{b!} \\\\ &= \exp(x)\cdot\exp(y) \end{align}

This can actually be done without writing a single sum. Consider the function $$ f(x, y) = \frac{e^x e^y}{e^{x+y}}. $$ Observe that $$ \frac{\partial f}{\partial x} = \frac{e^x e^y e^{x+y} - e^x e^y e^{x+y}}{(e^{x+y})^2} = 0. $$ Similarly, $$ \frac{\partial f}{\partial y} = \frac{e^x e^y e^{x+y} - e^x e^y e^{x+y}}{(e^{x+y})^2} = 0. $$ This shows that $f$ is a constant function. Now, we need only to use the series definition to show $f(0, 0) = 1$. Then, by rearrangement, we have the desired result: $$ e^{x+y} = e^x e^y. $$

My solution

Let $x,y \in \mathbb R$ and

$f(z) := \sum_{n=0}^\infty \left(\frac {x^n}{n!} \right )z^n$ and $g(z) := \sum_{n=0}^\infty \left(\frac {y^n}{n!} \right )z^n$. Then $\exp(x) \exp(y) = f(1)g(1)$. That is $$ f(z)g(z) = \sum_{n=0}^\infty \left( \sum_{k=0}^m \frac {x^m y^{n-m}}{m! (n-m)!} \right)z^n $$ $$ = \sum_{n=0}^\infty \frac 1 {n!} (x+y)^n z^n $$ thus $f(1)g(1) = \exp(x+y)$.