Intuition to why the Taylor series definition of $e^x$ matches with the algebraic definition of exponentiation with a rational $x$?

Question

For a rational $x = \frac{p}{q}$ where $p,q$ are integers and $q \neq 0$, $e^x$ can be defined in an algebraic way as follows: multiply $e$ repeatedly $p$ times and take the $q$-th root of the product.

On the other hand, $e^x$ is defined as sum of the following infinite series (which exists for all $x \in \mathbb{R}$)

$$e^x = 1 + \frac{x}{1!}+ \frac{x^2}{2!}+ \ldots$$.

I am interested to know if there is an intuition as to why these two definitions match at $x \in \mathbb{Q}$!!

There is also the following equivalent definition of $e^x$ $$e^x = \lim_{n \rightarrow \infty}\big(1 + \frac{x}{n}\big)^n$$.

Any intuition as to why these two definition match with the algebraic definition fo $e^x$ at rational points will be appreciated :)

Answer 1

Summarizing the comments of @MartinR and @IvanKaznacheyeu here:

Let $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots $. Then show that $e^x \cdot e^y = e^{x+y}$ for any $x,y \in \mathbb{R}$.

Define $e := e^1 = 1 + \frac{1}{1!} + \frac{1}{2!} + \ldots $. Now for any positive integer $p$, $$e^p = e^{1 + (p-1)} = e\cdot e^{p-1} (\ \because e^{x+y} = e^x\cdot e^y\ ) \ .$$ Unfolding the recurrence, we get $$e^p = \underbrace{e \cdot e \cdot \ldots \cdot e}_{p \text{ times}}$$

which is the algebraic definition of $e^p$.

Similarly for any integer $q \neq 0$, $$\underbrace{e^{1/q} \cdot e^{1/q} \cdot \ldots \cdot e^{1/q}}_{q \text{ times}} = e^{\underbrace{1/q + 1/q + \ldots + 1/q}_{q \text{ times}}} = e$$.

The proof will work for any definition of $e^x$ that gives the correct value for $e^1$ and satisfies $e^{x+y} = e^x \cdot e^y$ for all $x,y \in \mathbb{R}$.

Answer 2

For $x \in \mathbb{R}$, $e^x$ is defined as the unique solution to the ODE $f'(x) = f(x)$, $f(0) = 1$. Existence is proved by the power series approach. Uniqueness can be proved by some manipulations with the ODE (or by citing a general ODE theorem).Let $S(t)y_0 = e^{t}y_0$ be the solution of the ode $f'(t) = f(t)$, $f(0) = y_0$ evaluated at time $t$. By the uniqueness of the solution to the ODE, we get the same result whether going time $s + t$ or by going first time $s$ and then additional time $t$. Hence $S(s + t)y_0 = S(t)S(s)y_0$, i.e. $e^{s + t}y_0 = e^{t}e^{s}y_0$, so using $y_0 = 1$ gives $e^{s + t} = e^{s}e^{t}$. Now by routine algebra, we can get $(e^t)^{p/q} = e^{tp/q}$ for all rational $p/q$.