Why do we need to prove $e^{u+v} = e^ue^v$? In this book I'm using the author seems to feel a need to prove
$e^{u+v} = e^ue^v$
By
$\ln(e^{u+v}) = u + v = \ln(e^u) + \ln(e^v) = \ln(e^u e^v)$
Hence $e^{u+v} = e^u e^v$
But we know from basic algebra that $x^{a+b} = x^ax^b$.
Earlier in the chapter the author says that you should not assume $e^x$ "is an ordinary power of a base e with exponent x."
This is both a math and pedagogy question then, why does he do that?
So 2 questions really


*

*Do we need to prove this for such a basic property?

*If we don't need to, then why does he do it?  Fun?  To make it memorable?  Establish more neural connections?  A case of wildly uncontrolled OCD?


Also I've always taken for granted the property that $x^{a+b} = x^a x^b$.  I take it as an axiom, but I actually don't know where that axiom is listed.
 A: The necessity of a proof depends very much on how each of those things is defined and the domain over which the variables may vary. For example, if $e^x$ is defined by the power series $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for all $x \in \mathbb{C}$, then there is a very real need to prove that $e^{x + y} = e^x e^y$, and the proof is non-trivial. On the other hand, if $e$ is some constant defined elsewhere and $x$ and $y$ are positive integers, then there's essentially nothing to prove.
I would also like to add that the law doesn't hold everywhere. If $A$ and $B$ are square matrices (or endomorphisms of a vector space) that do not commute, then in general $\exp(A + B) \ne \exp(A) \exp(B)$. This fact allows us to study certain non-commutative groups using the tools of analysis and differential geometry — this is the field of study known as Lie theory.
A: One advanced reason for giving a proof is that in some related contexts such a formula breaks down. On the 2-adic numbers, the power series $A(x) = \exp(2x^2 - 2x)$ when expanded out and written in standard form turns out to converge on ${\mathbf Z}_2$, so in particular $A(1)$ is defined. One can prove $A(1)^2 = 1$, but although naively one may expect $A(1) = 1$ by just plugging 1 directly into the original definition I gave for $A(x)$, in fact $A(1) = -1$.  (The point here is that the naive calculation $A(1) = \exp(2 - 2) = \exp(2)\exp(-2) = 1$ is wrong since $\exp(2)$ doesn't make sense 2-adically.)
A: Mathematicians have a habit of using the same notation to denote many different concepts.  To justify their overloading, they tend to point to similarities between properties of those concepts.  You may think of the exponential as one concept, but it is actually a large family of related concepts:


*

*The exponentials $x^n$ where $x$ is an element of an arbitrary monoid and $n$ a non-negative integer,

*The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ an integer,

*The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ a rational (which are not guaranteed to exist in general, and are also not guaranteed to be unique in general),

*The exponential $x^n$ where $x$ is an element of a topological group where $x^n$ for $n$ rational is unique, and $n$ a real number (defined by limits as in Rudin),

*The exponential $e^x = 1 + x + \frac{x^2}{2} + ... $ where $x$ is an element of a topological ring containing $\mathbb{Q}$ and the series converges,

*etc.


It is easy to be fooled into thinking that these are all the same concept because they define essentially the same operation on $\mathbb{R}$, but 1) this is not obvious and requires proof, and 2) they generalize in different directions, so should be regarded as different in full generality.  Note that the condition for $x^n$ to be defined where $x$ is an element of a topological group and $n$ a real number is quite strong and rarely satisfied.  
Note also that your proof using logarithms runs into unnecessary subtleties when $u, v$ are taken to be complex numbers, since in this setting the logarithm is not uniquely defined.  Nevertheless, the exponential law still holds in this setting; it is one of the basic properties that mathematicians point to to justify calling something an exponential in the first place.
In higher mathematics, the exponential further generalizes to:


*

*The exponential map in Lie theory,

*The exponential map in Riemannian geometry,

*The exponential map in category theory,

*etc.


So it is good to keep in mind that "exponential" does not just refer to one thing.
A: In fact, such a proof is often necessary, which is why many authors write the function $e^x$ as $\exp(x)$ until they establish that it's just a "normal" exponent. For instance, if the original definition is given as
$$\exp(x) = \lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n,$$
then proving that $\exp(x + y) = \exp(x) \exp(y)$ is non-obvious, and certainly necessary.
A: One definition of $e^x$ is $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$. From this definition, it doesn't automatically follow that $e^x e^y = e^{x+y}$.
In fact, it doesn't even follow immediately that $e^x = \displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = (\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n)^x = (e)^x$. What this means is $e^x$ is just a short hand notation for the limit which after some analysis we realize it as $(e)^x$.
By limit arguments, we can now show that $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$, $\forall x \in \mathbb{R}$.
Now $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!})$.
Now we need to realize that we can rearrange the terms in the series and multiply terms of the two series since both of them converge absolutely.
Hence $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!}) = 1 + (x+y) + (\frac{x^2 + 2xy + y^2}{2!}) + (\frac{x^3 + 3x^2y + 3xy^2 + y^3}{3!}) + \cdots$
Now make use of the binomial theorem to get 
$$e^{x} e^{y} = e^{x+y}$$
PS: Though I have taken care to make sure the line of thought is right, you need to be careful when writing down the argument as to when you can interchange terms in an infinite series, multiply out two infinite series etc etc.
A: This may seem like a pretty pointless proof, at least on the surface, but I suspect there's some subtlety in the way the author's defined things here. (It may even appear circular at first, considering that the logarithm is often introduced as the inverse of the exponential function. Saying that, it can be derived the other way round, and this can sometimes be enlightening.)
A rigorous proof for integer exponents is very straightforward indeed, and follows simply from the definition of the exponential function (of arbitrary base). For arbitrary exponents, things get slightly more complicated. I present a more complete proof below.
So, let us suppose that the author began by defining the (natural) logarithm function,
$$\ln a = \int_1^a \frac{dx}{x} .$$
We can then prove the addition property of logarithms, $\ln (ab) = \ln a + \ln b$, by considering
$$
\ln (ab) 
= \int_1^{ab} \frac{1}{x} \; dx 
= \int_1^a \frac{1}{x} \; dx \; + \int_a^{ab} \frac{1}{x} \; dx 
=\int_1^{a} \frac{1}{x} \; dx \; + \int_1^{b} \frac{1}{t} \; dt 
= \ln (a) + \ln (b)
$$
(See this Wikipedia page for reference.)
The exponential function can of course be defined as the inverse of the logarithm, i.e.
$$exp(ln(a)) = a$$
Now, to prove the property of exponentials, $e^{u+v} = e^u e^v$, we start as follows.
$$\text{Let}\ u = \ln a, v = \ln b .$$
Then, using this property of logarithms and the definition of the inverse, consider
$$e^{u+v} = e^{\ln a + \ln b} = e^{\ln (ab)} = ab = e^u e^v .\ \square$$
That should hopefuly be straightforward enough to follow. There are of course other equivalent definitions of $exp$ and $ln$. (You can for example define the Taylor series of $exp$, use the Cauchy product, and then simplify, but that's slightly trickier.)
