The value of $\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{7}\right)\dots \left(1-\frac{1}{100}\right)$ is

1. Simplify each individual fraction in the product:
$$\left(\frac{4}{5}\right)\times \left(\frac{5}{6}\right)\times \left(\frac{6}{7}\right)\times \cdots \times \left(\frac{99}{100}\right)$$
This is a telescoping product where the denominator of each fraction cancels out the numerator of the subsequent fraction:
$$\frac{4}{\cancel{5}}\times \frac{\cancel{5}}{\cancel{6}}\times \frac{\cancel{6}}{\cancel{7}}\times \cdots \times \frac{\cancel{99}}{100}$$
After all cancellations, only the first numerator and the last denominator remain:
$$\text{Result}=\frac{4}{100}=\frac{1}{25}$$