1. The problem asks to prove a statement involving nonnegative integers $n$ and $r$. 2. Since the exact statement to prove is missing, let's assume it involves a common combinatorial identity such as the binomial coefficient identity: $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ 3. To prove this, we use the definition of factorial and properties of combinations. 4. For nonnegative integers $n$ and $r$ with $0 \leq r \leq n$, the number of ways to choose $r$ elements from $n$ elements is given by: $$\binom{n}{r} = \frac{n \times (n-1) \times \cdots \times (n-r+1)}{r \times (r-1) \times \cdots \times 1}$$ 5. This can be rewritten using factorials as: $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ 6. This formula counts the number of combinations without regard to order. 7. If the problem involves proving a specific identity, please provide the exact statement for a detailed proof.