Empty Range in Multiplication

Question

Why is the following true?

$$\prod_{i=6}^2 3^i = 1$$

I can't quite wrap my head around it. I understand that

$$\prod_{i=1}^k 3^i = \underbrace{3^1 \cdot 3^2 \cdots 3^{k-1} \cdot 3^k}_{3^{\sum_{j=1}^k}}$$

Is it because that sum is now zero and $3^0 = 1$?

Answer 1

Presumably, the author implicitly used a definition such that

$$\prod_{k=a}^b$$ involves all $k$ in $$a\le k\le b,$$ and if that range is empty, the product is $1$ by default (just as a sum is $0$ by default).

Other authors allow bounds reversal.

Answer 2

Well, given a ring $R$. By definition, the empty sum is equal to the zero element $0$ of the ring, and the empty product is equal to the unit element $1$ of the ring.