TITLE: Why does the Taylor expansion of $e^x$ satisfy exponential properties?
QUESTION [5 upvotes]: Suppose I knew nothing about the function $e^x$. If I wanted to find a power series that was its own derivative, I would logically start with the constant term, and first start by setting it to $1$. Then, the next term should be the antiderivative of the first term, giving me $x$. Doing this again would give me $\frac{x^2}2$. Repeating this process over and over again, I would get
$$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
Graphing a few terms of this, I might notice that this looks more and more like an exponential the more terms I graph. If I prove that this function satisfies the exponential relationship $f(x+y) = f(x)f(y)$, I would be able to prove that this series is an exponential function. How would I prove this? After this, how would I prove that the base of this exponential function is $e$, which is defined as $\displaystyle{\lim_{n \to \infty}} (1+\frac1n)^n$?
Edit: After expanding $(1+x)(1+y)$ and $(1+x+\frac{x^2}2)(1+y+\frac{y^2}2)$, I can see how extra terms get taken care of when the next degree is added. However, my second question still stands.

REPLY [2 votes]: THIS ARTICLE discusses six independent and equivalent characterizations of exponential function, $e^x$.


The OP assumes that the exponential function is characterized as the function $f(x)$ that satisfies the ordinary differential equation
$$f'(x)=f(x)\tag 1$$
subject to the initial condition $f(0)=1$.
Inasmuch as $f'(x)=f(x)$, $\forall x$, then by induction $f\in C^\infty$ with $f^{(n)}(x)=f(x)$.  Hence, $f^{(n)}(0)=f(0)=1$ and the Taylor series for $f(x)$ is given by
$$f(x)=\sum_{k=0}^\infty \frac{x^n}{n!}\tag2$$
This establishes the equivalence of the solution of the ODE $(1)$ and the Taylor series representation $(2)$.
That is to say, if we name the function $f(x)$ that satisfies $(1)$, subject to $f(0)=1$, the exponential function, then the exponential function is represented by the Taylor series given in $(2)$.  Note that the converse is also true.

Next, the OP tacitly assumes that the exponential function is characterized by the functional equation
$$f(x+y)=f(x)f(y)\tag 3$$
And that if the Taylor series in $(2)$ satisfies $(3)$, then the Taylor series representation is the exponential function.
Other solutions presented on this page have already established (e.g., use of Cauchy Product) that $f(x)$ as given by $(2)$ satisfies the functional equation $(3)$.


However, we have not shown that $f(x)$ as characterized by $(1)$, or $(2)$, or $(3)$ is the function $e^x$, where $e$ is defined as
$$e=\lim_{n\to \infty}\left(1+\frac1n\right)^n$$

To do so, one first needs to prove that
$$\lim_{n\to \infty}\left(1+\frac xn\right)^n=\left(\lim_{n\to \infty}\left(1+\frac1n\right)^n\right)^x =e^x\tag4 $$
If $x\in \mathbb{Q}$, then proof of $(4)$ is straightforward.  Then, by exploiting the density of the rational numbers, one can prove that $(4)$ is true for $x\in \mathbb{R}$.

Note that $(4)$ provides yet another unique characterization of the exponential function, which can be shown equivalent to $(1)$, $(2)$, and $(3)$.