1. **Stating the problem:** We want to prove or verify the inequality $$3^n < (n+1)!$$ for natural numbers $n \geq 4$. 2. **Recall definitions and formulas:** - The factorial function $(n+1)! = (n+1) \times n \times (n-1) \times \cdots \times 1$. - The exponential function $3^n$ means multiplying 3 by itself $n$ times. 3. **Check the base case $n=4$:** - Calculate $3^4 = 81$. - Calculate $(4+1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. - Since $81 < 120$, the inequality holds for $n=4$. 4. **Inductive step:** Assume the inequality holds for some $n=k \geq 4$, i.e., $$3^k < (k+1)!$$ 5. **Show it holds for $n=k+1$:** - Multiply both sides of the assumption by 3: $$3^{k+1} = 3 \times 3^k < 3 \times (k+1)!$$ - We want to prove: $$3^{k+1} < (k+2)! = (k+2) \times (k+1)!$$ - Since $k+2 > 3$ for $k \geq 4$, it follows that: $$3 \times (k+1)! < (k+2) \times (k+1)!$$ - Therefore: $$3^{k+1} < (k+2)!$$ 6. **Conclusion:** By mathematical induction, the inequality $$3^n < (n+1)!$$ holds for all natural numbers $n \geq 4$. **Final answer:** The inequality is true for all $n \geq 4$.