To find the least possible value of \(n\), we need \(f'(x) \geq 0\) for \(x > n\) because the function is increasing.

First, solve \(f'(x) = x^2 - 6x + 8 = 0\).

Factor the quadratic: \((x - 2)(x - 4) = 0\).

The roots are \(x = 2\) and \(x = 4\).

The quadratic \(x^2 - 6x + 8\) is positive for \(x < 2\) and \(x > 4\).

Since the function is increasing for \(x > n\), the least integer \(n\) such that \(f'(x) \geq 0\) is \(n = 4\).