1. **State the problem:** Simplify the expression $$\frac{n!}{(n-4)!}$$ assuming $n$ is an integer and $n \geq 4$. 2. **Recall the factorial definition:** For any integer $k \geq 0$, $$k! = k \times (k-1) \times (k-2) \times \cdots \times 1.$$ 3. **Rewrite the factorial in the numerator:** $$n! = n \times (n-1) \times (n-2) \times (n-3) \times (n-4)!$$ 4. **Substitute into the original expression:** $$\frac{n!}{(n-4)!} = \frac{n \times (n-1) \times (n-2) \times (n-3) \times (n-4)!}{(n-4)!}$$ 5. **Cancel the common factor $(n-4)!$ in numerator and denominator:** $$= \frac{n \times (n-1) \times (n-2) \times (n-3) \times \cancel{(n-4)!}}{\cancel{(n-4)!}}$$ 6. **Simplify the expression:** $$= n \times (n-1) \times (n-2) \times (n-3)$$ **Final answer:** $$\frac{n!}{(n-4)!} = n (n-1) (n-2) (n-3)$$