Show that exp(x) is of the form $e^x$? Let me define the function $\exp(x)$ as that which satisfies:
$$ \frac{d}{dx}\exp(x) = \exp(x) $$
We can show that 
$$ \exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$
So far so good, we have just defined the exponential by the differential equation it satisfies.
How can we show that the exponential, as defined above, is of the form:
$$\exp(x) = e^x,$$
where $e$ is a real number.
In other words, how can we show, for example that:
$$ e^a = \prod_{m=1}^{a}e^1$$
$$\sum_{n=0}^\infty \frac{a^n}{n!} = \prod_{m=1}^a \left[ \sum_{l=0}^{\infty} \frac{1}{n!} \right]$$
for any positive integer $a$.
 A: First note that apart from $\exp'=\exp$, you need to postulate $\exp(0)=1$. Then it is the uniqueness of the solution to the differential equation that determines the desired properties:
The derivative of $f(x)= b\cdot \exp(a+x)+c\exp(x)$ apparently is $$f'(x)=b\exp'(a+x)+c\exp'(x)=b\exp(a+x)+c\exp(x)=f(x).$$ So if we adjust $b,c$ to ensure $f(0)=1$, we conclude that $f\equiv \exp$.
If $\exp(a)=0$, we take $b=c=1$ and infer $\exp(x)=\exp(a+x)+\exp(x)$ for all $x$, i.e., $\exp\equiv 0$, contradiction.
Hence we may assume $\exp(a)\ne 0$, can take $c=0$ and $b=\frac 1{\exp(a)}$, and find
$$ \exp(x)=\frac{\exp(a+x)}{\exp(a)}$$
for all $a,x\in\Bbb R$. In other words,
$$ \exp(a+b)=\exp(a)\exp(b)$$
for all $a,b\in\Bbb R$.
From this, we learn among others (by induction) that $$\exp(n)=\exp(1)^n$$ for all $n\in\Bbb Z$.
A: It suffices to show that $$\exp(x+y)=\exp (x)\exp(y)$$But this can be easily shown since $$\exp(x)\exp(y){=\sum_{n=0}^{\infty}{x^n\over n!}\sum_{m=0}^{\infty}{y^m\over m!}\\=\sum_{n,m=0}^{\infty}{x^ny^m\over n!m!}\\=\sum_{k=0}^\infty{1\over k!}\sum_{m+n=k}{k!x^ny^{k-n}\over n!(k-n)!}\\=\sum_{k=0}^\infty {(x+y)^k\over k!}\\=\exp(x+y)}$$
A: One way is through $$e^x\equiv\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n 
$$
You can show through the binomial expansion that this recipe gives the Taylor series for $\exp x$ and thence that it solves the differential equation; or by differentiating the expression with 'lim' in it directly; and also showing that this expression is also the natural extension of exponentiation into the domain of reals from its definition for integer argument in terms of iterated multiplication. If you ply these various relations backward & forward in every conceivable direction, bringing in also the fundamental definition of differentiation in terms of limits, it becomes clear that it all constitutes a beautifully internally consistent system of relations. And you can also get 
$$\frac{dy}{dx}\equiv a\frac{y}{x}$$ & 
$$x=1\implies y=1$$
as a definition of $x^a$ for general real $a$.
