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
$$
e_n = \sum_{m=0}^n c_md_{n-m}
$$
 A: \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}
A: $$
\begin{align*}
& \exp(x+y)=\sum_{n=0}^{\infty}\frac{1}{n!}(x+y)^n = \sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{(n-k)!k!}
x^ky^{n-k} \\
=& \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} = \lim_{N \to \infty}\sum_{n=0}^{N}\sum_{k=0}^{n} \frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}\\
& \mathrm{\quad make \; a \; bijective \;map\; between} \left\{(n,k):0 \le n \le N,0 \le k\le n\right\} \\&\mathrm{\; and } \left\{(n,k):0 \le k \le N,k \le n\le N\right\} \\
=& \lim_{N \to \infty}\sum_{n=0}^{N}\sum_{k=0}^{n} \frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}= \lim_{N \to \infty}\sum_{k=0}^{N}\sum_{n=k}^{N}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} \\
= &\lim_{N \to \infty} \sum_{k=0}^{N}\frac{x^k}{k!}\sum_{n=k}^{N}\frac{y^{n-k}}{(n-k)!} = \lim_{N \to \infty} \sum_{k=0}^{N}\frac{x^k}{k!}\sum_{n=0}^{N-k}\frac{y^{n}}{n!} = \sum_{k=0}^{\infty}\frac{x^k}{k!}\sum_{n=0}^{\infty}\frac{y^{n}}{n!} \\
=&\sum_{k=0}^{\infty}\frac{x^k}{k!}\sum_{n=0}^{\infty}\frac{y^{n}}{n!}=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\frac{x^k}{k!}\frac{y^{n}}{n!} = \exp(x)\exp(y)
\end{align*} 
$$
A: 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: $$ e^{x+y} = e^x e^y. $$
A: 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 $f(1)g(1) = \exp(x+y)$.
A: $A(t)=\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}t^n$
$B(t)=\exp(y) = \sum_{n=0}^\infty \frac {y^n}{n!}t^n$
$C(t) = A(t)*B(t)=\sum_{n=0}^\infty (\sum_{k+z=n}^\ \frac {x^k}{k!}*\frac {y^z}{z!})t^n=\sum_{n=0}^\infty \frac {(x+y)^n}{n!}t^n=exp(x+y)$
and use $t=1$
sry i was too late^^
A: Euler formula says that (exponential form of a complex number) $$e^{i\theta}=\cos\theta+i\sin\theta.$$ Therefor $$e^{x+y}=\cos(-i(x+y))+i\sin(-i(x+y))\\=\cos(xi+yi)-i\sin(xi+yi)\\=(\cos ix\cos iy-\sin ix\sin iy)+i(\sin ix\cos iy+\cos ix\sin iy)\\=(\cos ix+i\sin iy)(\cos iy+i\sin iy)\\=e^xe^y.$$
A: 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.
