How to prove that $\exp(x_1+x_2)=\exp(x_1)\exp(x_2)$? In my analysis class, we covered the properties of the exponential function (as of now we use $\exp$ instead of $e$). One of the properties of $e$ is that $\exp(x_1+x_2)=\exp(x_1)\exp(x_2)$. In high school this was just assumed knowledge but now we have to prove that this statement is indeed true, which is proving to be quite difficult. 
I assume that one would need to work out the derivatives of both side and see if they are equal. Then if $f(0)=1$ we can assume that both sides of the equation are true. 
 A: From your comments it appears that you are using the definition of $\exp(x) $ as the unique solution to $f'(x) =f(x), f(0)=1$. It can be proved that such a solution must be non-zero for all $x$ (see second part of this answer). Thus $\exp(x) \neq 0$ for all $x$. Let $a$ be any arbitrary real number and consider the function $g$ defined by $g(x) =\exp(x+a) /\exp(x) $. We have $$g'(x) =\frac{\exp (x) \exp(x+a) - \exp(x+a) \exp(x) } {(\exp(x)) ^2}=0$$ and thus $g$ is a constant. It follows that $g(x) =g(0)=\exp(a)$ and thus $$\exp(x+a) =\exp(x) \exp(a) $$ Now replace $x$ by $x_1$ and $a$ by $x_2$.

The above technique does not work if one chooses $$g(x) =\exp(x+a) - \exp(x) \exp(a) $$ because one can't see that $g'(x) =0$ in very obvious manner. But this can be done with some effort. Using above definition of $g$ one gets $g'(x) =g(x),g(0)=0$. Ideally when one studies the definition of exponential function as a solution of differential equation $f'(x) =f(x), f(0)=1$ then the first step is to show that if the solution exists then it must be unique. And that uniqueness is shown by the following:

Theorem: If function $g:\mathbb{R} \to\mathbb{R} $ satisfies $g'(x) =g(x), g(0)=0$ then $g(x) =0$ for all values of $x$.

On the contrary assume that there is some $a$ such that $g(a) \neq 0$ and consider $$h(x) =g(a+x) g(a-x) $$ then $$h'(x) =g(a+x) g(a-x) - g(a+x) g(a-x) =0$$ so that $h$ is constant and $h(x) =h(0)=g(a)g(a)>0$ or in other words $g(a+x) g(a-x) >0$. Putting $x=a$ and noting that $g(0)=0$ we see that $0>0$ and this contradiction shows that $g(x) =0$ for all $x$ and we are done. The same proof can be used to show that $\exp(x) \neq 0$. We just need to consider $h(x) =\exp(x) \exp(-x) $. 
A: if you are familiar with the Cauchy-product $CP$ you can proof this like follows:
$$exp(z) \cdot exp(w)\sum_{n=0}^{\infty} \frac{z^n}{n!}\sum_{n=0}^{\infty} \frac{w^n}{n!} \overset{CP}{=}\\ \sum_{n=0}^{\infty} \left(\sum_{k=0}^{\infty} \frac{z^k}{k!} \frac{w^{n-k}}{(n-k)!} \right)=\sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^{\infty} \frac{n!}{k!(n-k)!}z^kw^{n-k} \right)= \\ \sum_{n=0}^{\infty}\left(\frac{1}{n!}\sum_{k=0}^{\infty}\binom{n}{k} z^k w^{n-k} \right) = \sum_{n=0}^{\infty} \frac{1}{n!}(z+w)^n = exp(z+w)$$
A: This is one of the basic properties of an exponential function. Assuming that you know the basic connection between it and natural logarithm, this is that $ e^x $ is the inverse of $ \ln{x} $. With that being said, we can write the following
$$ \ln{e^{x+y}} = x + y = \ln{e^x} + \ln{e^y} = \ln{e^x \cdot e^y}$$
Ultimately, since natural logarithm and exponential function are fundamentally related (trivially speaking one-to-one), we conclude that
$$ e^{x+y} = e^x \cdot e^y $$
Be aware that such proof does not use derivatives. I hope that helps.
