Okay, let $y\in \mathbb{R}\;$ be fixed and consider the function $\;f(x)=e^{x+y}$.
How do I show that $\,f\,$ is analytic in $\,\mathbb{R}\,$ and use this fact and the Taylor series in $\,x=0\,$ of this function to derive the following formula?
$$e^{x+y}=e^x\,e^y\qquad x,y\in \mathbb{R}$$
Help appreciated!!
Start by computing $f^{(n)}(0)$ for each $n$ and showing the error term goes to zero.
In this case, since the derivatives are nonnegative, you can use Bernstein's criterion: If $f$ and each of its derivatives is nonnegative over a closed interval $[0,r]$ then for $x\in[0,r)$ the Taylor series $$\sum_{k\geq 0}\frac{f^{(k)}(0)}{k!}x^k$$ converges to $f(x)$. Of course in this case you can take $r$ as large as you want.
You should get that $f^{(n)}(0)=e^y$ for each $n$. This means that $$f=\sum_{k\geq 0}\frac{e^y}{n!}x^n$$
or $$f=e^y\sum_{k\geq 0}\frac{x^n}{n!}$$
Then observe that $$e^x=\sum_{k\geq 0}\frac{x^n}{n!}$$
over the real line.