For writing a (german) article about the power with natural degree I have the following question:
In school one defines the power with natural degree via
$$n^k = \underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}$$
In calculus normally recursion is used to introduce the power:
$$\begin{align} n^0 & := 1 \\ n^{k+1} & := n \cdot n^k \end{align}$$
Question: What are the disadvantages of the expression $\underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}$ and why it is not used to define the power in advanced mathematics (although it is intuitive)?
Note: I'm sorry, I had an ambiguous question before. I do not want to know, why we define the power. I want to know, why we use recursion and not the expression $\underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}$ to define the power in advanced mathematics...
My ideas so far:
- $\underbrace{\ldots}_{k \text{ times}}$ is no operator, which was introduced before.
- definitions with recursion lead naturally to a scheme, how properties of these concepts can be proved via induction
$n^k = \underbrace{n\cdot n \cdots n}_{k\ \textrm{times}}$ only works when $k$ is a positive integer. In order to define exponentiation more generally, we must refine the definition of exponentiation several times:
For these reasons, the "repeated multiplication" definition fails for almost every interesting value of $k$. The goal is to find a definition that reduces to "repeated multiplication" in the special case that $k \in \mathbb{Z}^+$, but that works more generally for all real numbers when the base is positive.
To address your edit, which is not something I've actually encountered, but I will comment nonetheless:
If you define $n^{k+1} = n \cdot n^k$, this definition is satisfactory for all values of $k$, presuming you have defined exponentiation rigorously. This differs from the repeated multiplication definition in that you may start from an arbitrary $k$. This is actually one step in the process of defining exponentiation. More generally, we wish to declare that $n^{k_1}n^{k_2} = n^{k_1+k_2}$ regardless of the classes of real number to which $k_1$ and $k_2$ belong. The "recursive" definition you presented is a special case of this.