70. How many members will result after expanding all the brackets in the expression: $(a+1)(b+1)(c+1)(d+1)(e+1)(f+1) \times$ $\times(g+1)$ ?
70. Each of the members that result from expanding the brackets in the expression

$$
\begin{aligned}
(a+1)(b+1)(c+1)(d+1) & (e+1) \times \\
& \times(f+1)(g+1)
\end{aligned}
$$

is a product of seven factors (since there are seven brackets). Any of these factors is either a letter or the digit 1. Thus, we need to find the number of products of seven factors, each of which can be in "two states." Therefore, our problem is equivalent to the following:

Find the number of ways to light up seven light bulbs, each of which can be on or off. Answer: $2^{7}$.