1. **State the problem:** We need to find the number of 4-digit numbers formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 such that each number contains the digit 1 at least once. 2. **Total 4-digit numbers from given digits:** The first digit can be any from 1 to 7 (cannot be 0 as it won't be a 4-digit number), so 7 choices. Each of the remaining three digits can be any of the 8 digits (0 to 7), so $8^3$ choices. Total numbers = $7 \times 8^3 = 7 \times 512 = 3584$. 3. **Numbers without digit 1:** To find numbers that do not contain digit 1, the first digit can be chosen from {2,3,4,5,6,7} (6 choices, excluding 0 and 1). Each of the remaining three digits can be chosen from {0,2,3,4,5,6,7} (7 choices, excluding 1). Total numbers without digit 1 = $6 \times 7^3 = 6 \times 343 = 2058$. 4. **Numbers with at least one digit 1:** Use complementary counting. Number with at least one 1 = Total numbers - Numbers without 1 = $3584 - 2058 = 1526$. **Final answer:** There are $\boxed{1526}$ such 4-digit numbers containing the digit 1 at least once.