In the following example in this MIT lecture, how did we establish that a $2^{n+1} \cdot 2^{n+1} = 4(2^n \cdot 2^n)$?

Question

In the following example at the minute 1:06:43 in this video: https://www.youtube.com/watch?v=z8HKWUWS-lA&t=1810s

While using proof by induction, to make use of the assumption, while validating the implication/inductive step, the 2^n+1 square is divided into 4 squares each is 2^n * 2^n. Now how is that established, how is (2^n+1 * 2^n+1) = 4(2^n * 2^n)?

Answer 1

You have

$$\begin{equation}\begin{aligned} (2^{n+1})(2^{n+1}) & = (2(2^{n}))(2(2^{n})) \\ & = 2(2^{n})2(2^{n}) \\ & = 4(2^{n})(2^{n}) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Answer 2

This boils down to exponent rules ($a^{m+n} = a^m a^n$) and commuting terms: \begin{align*} \left(2^{n+1} \cdot 2^{n+1}\right) & = \left((2^n)(2) \cdot (2^n)(2) \right) \\ & = 4(2^n \cdot 2^n) \end{align*}