72. Find the product:

$$
\left(1-\frac{1}{4}\right) \cdot\left(1-\frac{1}{9}\right) \cdot\left(1-\frac{1}{16}\right) \ldots\left(1-\frac{1}{225}\right)
$$
72. $\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \ldots\left(1-\frac{1}{225}\right)=$ $=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \ldots \cdot \frac{224}{225}=\frac{1 \cdot 3}{2^{2}} \cdot \frac{2 \cdot 4}{3^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdot \ldots \cdot \frac{14 \cdot 16}{15^{2}}$.

By performing the sequential cancellations, we get the first factor $\frac{1}{2}$, the last factor $\frac{16}{15}$, and all intermediate factors are ones. Therefore, the desired product is $\frac{1}{2} \cdot \frac{16}{15}=\frac{8}{15}$.