prove $s(x+y)=s(x)s(y)$ I am asked to prove the following:

Let $s(x):=\sum_{n=0}^{\infty}\binom{x}{n}$. Then $s(x+y)=s(x)s(y)$.

I don't know how to start. I am thinking about $\exp(x)$ function with $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$. Am I okay with this start? I would then begin like this: 
$$\sum_{n=0}^{n}\dfrac{x^n}{n!} \cdot \dfrac{y^n}{n!} = .. help = \frac{1}{n!}(x+y)^n$$
I am not sure how to continue or whether I am okay. 
Thanks for help.
 A: $$\exp(x) \exp(y) = \left(\sum_{m=0}^{\infty} \dfrac{x^m}{m!}\right)\left(\sum_{n=0}^{\infty} \dfrac{y^n}{n!}\right) = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \dfrac{x^m}{m!}\dfrac{y^n}{n!}\\ = \sum_{k=0}^{\infty} \sum_{m+n=k} \dfrac{x^m}{m!}\dfrac{y^n}{n!} = \sum_{k=0}^{\infty} \sum_{n=0}^k \dfrac{x^{k-n}}{(k-n)!}\dfrac{y^n}{n!}$$
Now note that
$$\dfrac{x^{k-n}}{(k-n)!}\dfrac{y^n}{n!} = \dfrac1{k!} \dbinom{k}n x^{k-n} y^n$$
Hence, we get that
$$\exp(x) \exp(y) = \underbrace{\sum_{k=0}^{\infty} \dfrac1{k!} \sum_{n=0}^k \dbinom{k}n x^{k-n} y^n = \sum_{k=0}^{\infty} \dfrac{(x+y)^k}{k!}}_{\text{Using binomial theorem}} = \exp(x+y)$$
EDIT
Answering the updated question. First note the following combinatorial identity.
$$\dbinom{x+y}k = \sum_{n=0}^k \dbinom{x}n \dbinom{y}{k-n} \,\,\,\,\,\,\,\,\,\,\,\, (\star)$$
Hence, $$s(x+y) = \sum_{k=0}^{\infty} \dbinom{x+y}k = \sum_{k=0}^{\infty}\sum_{n=0}^k \dbinom{x}n \dbinom{y}{k-n} = \sum_{k=0}^{\infty}\sum_{m+n=k} \dbinom{x}n \dbinom{y}{m}\\ = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \dbinom{x}n \dbinom{y}{m} = \left(\sum_{n=0}^{\infty} \dbinom{x}n\right) \left(\sum_{m=0}^{\infty} \dbinom{y}{m} \right) = s(x) s(y)$$
If $x,y \in \mathbb{N}$, there is a nice combinatorial proof for $(\star)$. Consider a bag with $x$ red balls and $y$ blue balls. We want to choose a total of $k$ balls from these two bags. Hence, the number of ways of doing this is $\dbinom{x+y}k$. We can also count this by another method. Choose $n$ red balls and $k-n$ blue balls. The number of ways is $\dbinom{x}n \dbinom{y}{k-n}$. Now $n$ can vary from $0$ to $k$. Hence, the total number of ways of choosing $k$ balls from both these bags is $\displaystyle \sum_{k=0}^n \dbinom{x}n \dbinom{y}{k-n}$. Hence, we get that $$\dbinom{x+y}k = \sum_{n=0}^k \dbinom{x}n \dbinom{y}{k-n}$$
A: We have the generalized binomial theorem
$$ (1+z)^x=  \sum_{n=0}^\infty \binom{x}{n}z^n $$
where $\binom{x}{n}$ denotes the generalized binomial coefficient
$$ \binom{x}{n} = \frac{x(x-1)(x-2)\cdot\ldots\cdot(x-n+1)}{n}$$
Provided $x > -1$, this series converges with radius of convergence $1$, and specifically for $z=1$; convergence at $z=1$ is absolute if and only if $x > 0$. Thus, assuming $x > 0$, we have $s(x)=(1+1)^x=2^x$, and now the problem is trivial:  $$s(x)s(y)=2^x2^y=2^{x+y}=s(x+y)$$ for $x,y>0$.
A: First you must observe that the series converges absolutely.  Then we can use the fact that $\sum a_{n}b_{n}=AB$ if $\sum a_{n}=A$ and $\sum b_{n}=B$ and at least one of the series is absolutely convergent.  With this, we compute:
\begin{align*}
s(x)s(y)
&=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\sum_{m=0}^{\infty}\frac{y^{n}}{n!}\\
&=\sum_{n=0}^{\infty}\sum_{j=0}^{n}\frac{x^{j}y^{n-j}}{j!(n-j)!}\\
&=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{j=0}^{n}\binom{n}{j}x^{j}y^{n-j}\\
&=\sum_{n=0}^{\infty}\frac{(x+y)^{n}}{n!}\\
&=s(x+y).
\end{align*}
A: Probably something like
$$\sum_{n=0}^\infty \frac{(x+y)^n}{n!}=\sum_{n=0}^\infty\sum_{i=0}^n \frac{1}{n!}\binom{n}{i}x^iy^{n-i}=\sum_{n=0}^\infty \sum_{i=0}^n \frac{x^i}{i!}\frac{y^{n-i}}{(n-i)!}=\left(\sum_{j=0}^\infty \frac{x^j}{j!}\right)\left(\sum_{k=0}^\infty \frac{y^k}{k!}\right).$$
