Use mathematical induction to prove that \(n^{2}

Here's a step-by-step explanation of the solution: Step 1: Base case We need to show that the inequality $n^2<n!$ holds true for all values of $n\ge 4$ . So, we start by checking if the statement is true for the base case $n=4$ : $4^2=16$ and $4!=4\times 3\times 2\times 1=24$ . Therefore, since $16<24$, the statement holds true for $n=4$. Step 2: Inductive hypothesis Now, we assume that the statement is true for an arbitrary $k\ge 4$ . In other words, we assume: $k^2<k!$ Step 3: Inductive step We want to prove that the inequality also holds for the value of $k+1$ . Hence, we need to prove that: $(k+1{)}^2<(k+1)!$ We start by expanding the left-hand side of the inequality: $(k+1{)}^2=k^2+2k+1$ Using the inductive hypothesis, we know that $k^2<k!$. Hence, we multiply both sides of the inductive hypothesis by $k+1$ : $k^2(k+1)<k!(k+1)$ We can now see that: $k^3+k^2>k^2+2k+1$ This means that the left-hand side of the inequality is less than the right-hand side: $k^2+2k+1<k^3+k^2<k!(k+1)$ Therefore, we have shown that $(k+1{)}^2<(k+1)!$ when $k^2<k!$. Step 4: Conclusion By mathematical induction, we have proved that the statement $n^2<n!$ holds true for all values of $n\ge 4$ .