Divisors

As $5040 = 2^4\times3^2\times5\times7$ , you can observe that $2, 4, 8$ is in $2^4$ , 6 is a result of $2\times3$ , and $5, 7, 9$ has the respective factor. This means 1 to 9 is all factor to 5040, and 10 is also included as $10 = 2\times 5$ . The smallest positive which is not a factor is then be 11, 13, 17 and so forth...

The divisors of 5040 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12 , 14, 15, 16, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 48, 56, 60, 63, 70, 72, 80, 84, 90, 105, 112, 120, 126, 140, 144, 168, 180...... and there is no 11

By the fundamental theorem of arithmetic, we know every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. Thus, the smallest integer that does not divides evenly into 5040 will be a prime number. The unique prime factors of 5040 = 2^4 * 3^2 * 5 * 7. The list of the first prime numbers are 2, 3, 5, 7, and 11... Thus the smallest integer is 11, which does not evenly divide into 5040.

7! =2^4×3^2×5×7. so we have 8,9,10 also the divisors but 11 is a prime no. and can't be generated by these numbers.