Prove $\exp(x+y) = \exp(x) \exp(y)$ for $\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$ I am trying to prove $\exp(x+y) = \exp(x) \exp(y)$.

I may use that $$\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$$
I further know how to multiply two power series in one point, i.e. if $f(x) = \sum_{n=0}^\infty c_n(x-a)^n$ and $g(x) = \sum_{k=0}^\infty d_n(x-a)^n$ then
$$
f(x)g(x) = \sum_{n=0}^\infty e_n(x-a)^n
$$
with
$$
e_n = \sum_{m=0}^n c_md_{n-m}
$$
\begin{align} \exp(x+y)&amp;=\sum_n\frac{(x+y)^n}{n!} \\\\  &amp;=\sum_{n}\frac{1}{n!}\sum_{a+b=n} {n \choose a} x^ay^b \\\\
&amp;= \sum_{n}\frac{n!}{n!}\sum_{a+b=n}\frac{x^a}{a!}\frac{y^b}{b!} \\\\
&amp;= \sum_{a,b} \frac{x^a}{a!}\frac{y^b}{b!} \\\\
&amp;= \exp(x)\cdot\exp(y)
\end{align}
Assume the function $\frac{exp(x+y)}{exp(x)}$, where $y$ is some constant. Differentiate with respect to $x$ and find that the derivative is zero. Therefore $\frac{exp(x+y)}{exp(x)}=c\in\mathbb{R}$. From the fact that $exp(0)=1$ we get $exp(y)=c$ which proves what was asked.