Example 3 Figure 4 is a "sheep's head" pattern, the method of which is: starting from square (1), using one of its sides as the hypotenuse, an isosceles right triangle is constructed outward, and then using its legs as sides, squares (2) and (2)' are constructed outward, ... and so on. If the side length of square (1) is $64 \mathrm{~cm}$, then the side length of square (7) is $\qquad$ $\mathrm{cm}$.
Solution: By the Pythagorean theorem, the side length of square (2) is $64 \times \frac{\sqrt{2}}{2}=32 \sqrt{2}(\mathrm{~cm})$,
the side length of square (3) is
$$
32 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{2}=32(\mathrm{~cm}) \text {, }
$$

the side length of square (4) is
$$
32 \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{3}=16 \sqrt{2}(\mathrm{~cm}) \text {, }
$$

the side length of square (5) is
$$
16 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{4}=16(\mathrm{~cm}) \text {, }
$$

the side length of square (6) is
$$
16 \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{5}=8 \sqrt{2}(\mathrm{~cm}) \text {, }
$$

the side length of square (7) is
$$
8 \sqrt{2} \times \frac{\sqrt{2}}{2}=64 \times\left(\frac{\sqrt{2}}{2}\right)^{6}=8(\mathrm{~cm}) \text {. }
$$

Investigation: The side length of square (ㅈ) is
$$
64 \times\left(\frac{\sqrt{2}}{2}\right)^{n} \mathrm{~cm} .
$$