Answer
See explanation
Work Step by Step
Below is a standard induction proof using the *recursive definition of product* that shows
\[
\prod_{i=1}^n (a_i\,b_i)
\;=\;
\Bigl(\prod_{i=1}^n a_i\Bigr)\,\Bigl(\prod_{i=1}^n b_i\Bigr)
\quad
\text{for all positive integers }n.
\]
---
## 1. Recursive Definition of Product
For a positive integer \(n\), we define
\[
\prod_{i=1}^n x_i
\;=\;
\begin{cases}
x_1, & \text{if }n=1,\\[6pt]
\Bigl(\prod_{i=1}^{n-1} x_i\Bigr)\,\cdot\,x_n, & \text{if }n>1.
\end{cases}
\]
---
## 2. Proof by Induction
### Base Case \((n=1)\)
When \(n=1\),
\[
\prod_{i=1}^1 (a_i b_i) = a_1 b_1,
\]
while
\[
\Bigl(\prod_{i=1}^1 a_i\Bigr)\,\Bigl(\prod_{i=1}^1 b_i\Bigr)
= a_1 \cdot b_1.
\]
Clearly these are equal. So the statement holds for \(n=1\).
### Inductive Step
Assume the statement is true for some positive integer \(n\). That is, assume
\[
\prod_{i=1}^n (a_i b_i)
\;=\;
\Bigl(\prod_{i=1}^n a_i\Bigr)\,\Bigl(\prod_{i=1}^n b_i\Bigr).
\]
We must prove it for \(n+1\). Using the recursive definition of product on \(\{(a_i b_i)\}\):
\[
\prod_{i=1}^{n+1} (a_i b_i)
\;=\;
\Bigl(\prod_{i=1}^{n} (a_i b_i)\Bigr)\;\cdot\;(a_{n+1} b_{n+1}).
\]
By the inductive hypothesis, the portion in parentheses is
\[
\prod_{i=1}^{n} (a_i b_i)
\;=\;
\Bigl(\prod_{i=1}^n a_i\Bigr)\,\Bigl(\prod_{i=1}^n b_i\Bigr).
\]
Hence
\[
\prod_{i=1}^{n+1} (a_i b_i)
\;=\;
\Bigl(\prod_{i=1}^n a_i\Bigr)\,\Bigl(\prod_{i=1}^n b_i\Bigr)\;\cdot\;(a_{n+1} b_{n+1})
\;=\;
\Bigl(\prod_{i=1}^n a_i \cdot a_{n+1}\Bigr)\,\Bigl(\prod_{i=1}^n b_i \cdot b_{n+1}\Bigr).
\]
Using the recursive definition again, this is exactly
\[
\Bigl(\prod_{i=1}^{n+1} a_i\Bigr)\,\Bigl(\prod_{i=1}^{n+1} b_i\Bigr).
\]
Thus the identity holds for \(n+1\). By mathematical induction, it is true for all positive integers \(n\).
---
## 3. Conclusion
We have shown, using the recursive definition of product and induction, that for all \(n \ge 1\),
\[
\boxed{\prod_{i=1}^n (a_i\,b_i)
\;=\;
\Bigl(\prod_{i=1}^n a_i\Bigr)\,\Bigl(\prod_{i=1}^n b_i\Bigr).}
\]
