Question

For which nonnegative integers n is n^2 ≤ n!? Prove your answer

Answer 1

The inequality
`n^2 \leq n!` is valid for all nonnegative integers n. For n=0 and n=1, both sides of the inequality are equal to 1, and for
`n\geq 1`, n! grows faster than
`n^2`, satisfying the inequality.

Step-by-step explanation:

The question asks for which nonnegative integers n is
`n^2` ≤ n! The factorial function n!, which means the product of all positive integers from 1 to n, grows very quickly. For any integer value of n ≥1, it is evident that n2 will always be less than or equal to n! because the latter includes n as a factor and has additional factors that increase its value beyond
`n^2`.

However, when n=0 or n=1,
`n^2` = n! since both equal 1. To prove this, we can use the fact that the factorial function is 1 when n=0 or n=1 and then increases more rapidly than the square of n, as each subsequent term in the factorial introduces an additional multiplicand that is greater than or equal to 2. Therefore, the condition
`n^2` ≤ n! is satisfied for all nonnegative integers n.