Is it wrong to "imagine" a one when thinking about exponentiation? (e.g. $3^2 = 1 \times 3 \times 3$)

Question

This might be a bit of a basic question, but I'm going through Khan Academy to refresh my math skills in order to pursue a self-study of higher mathematics, so I'm really focused on the "why" of the rules.

One thing that has been bothering me is that I've been imagining a $1$ in front of the expanded form of an exponentiation. I haven't seen it described anywhere, but it's the only way it makes logical sense to me, so I wanted to check that I wasn't making some huge error.

I think the most obvious example of this is for fractional bases with negative exponents, since I need to "force" the reciprocal of the first iteration of the base.

Take $\frac{2}{3}^{-3}$ as an example. Without the imaginary $1$, my brain wants to evaluate it like this

$$\frac{2}{3}^{-3} = \frac{2}{3} \div \frac{2}{3} \div \frac{2}{3} = \frac{2}{3} \cdot \frac{3}{2} \cdot \frac{3}{2} = \frac{18}{12} = 1\frac{1}{2}$$

This is obviously wrong.

If I add the imaginary $1$, everything works out fine:

$$\frac{2}{3}^{-3} = 1 \div \frac{2}{3} \div \frac{2}{3} \div \frac{2}{3} = \frac{1}{1} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} = \frac{27}{8} = 3\frac{3}{8}$$

Am I missing something terribly obvious, or is this a common way to look at the problem (or an uncommon way to describe a common understanding)? I'm assuming that the concept of $n^0 = 1$ will play into this somewhere.

Answer 1

One nice thing about an 'imaginary' $1$ before an exponent expression is it makes it obvious why, for instance, $2^0=1$ ($2^0=1$, with $0$ following factors of $2$, but the $1$ is still there).

This works so well because $1$ is the multiplicative identity--it's like a blank slate for multiplication. Whatever you multiply by next, the result is itself (what you multiplied by). It also is a way of seeing why $0! = 1$.

Answer 2

Basically, you can always multiply anything by $1$ without changing it's value. This is what we do when extending fractions, for example: $$\frac12 = \frac12 \cdot \frac22 = \frac24$$ However, you main problem seems to be the definition of $$x^{-k} = \frac1{x^k}$$ If you write this correctly with divison, you end up with $$\left(\frac23\right)^{-3} = \frac1{\frac23} \cdot \frac1{\frac23} \cdot \frac1{\frac23} = \frac32\cdot\frac32\cdot\frac32 = \frac{27}8$$ It is simply not true that $x^{-k} = \underbrace{x\div x \div \ldots \div x}_{k \text{ times}}$