How do I prove the statement below by the method of mathematical induction?

Question

If $n ∈ N$ and $n \ge 4$ , how do I prove by the method of induction that $n! > n^2$?

Answer 1

"Proof by induction" is pretty "cut and dried". First you prove the statement for the smallest value of n- here n= 4. 4!= 4(3)(2)(1)= 24. $4^2= 16$. 24> 16 so that's done.

Now, assume the statement is true for some integer "k"- that $k!> k^2$- and prove from that that it is true for "k+ 1". $(k+1)!= (k+ 1)(k!)> (k+ 1)(k^2)= k^3+ k^2$. Now, $(k+ 1)^2= k^2+ 2k+ 1$ The problem reduces to showing that, for $k\ge 4$, $k^3+ k^2\le k^2+ 2k+ 1$ or, canceling the "$k^2$" terms, $k^3\ge 2k+ 1$.

Answer 2

We must first prove the base case, that $4! > 4^2$ which is true, since $24 > 16$.

Now, assume that $n! > n^2$. We will prove that $(n+1)! > (n+1)^2$.Now, we notice that $(n+1)! = (n+1) \cdot n!$, so we can divide $n+1$ on both sides, so that we want to prove that $n! > n+1$. However, we know that $n! > n^2$, so we just need to prove that $n^2 > n+1$, for $n \geq 4$. Therefore, we need to prove that $$n^2 - n - 1 > 0 \implies n^2 - n + \frac{1}{4} - \frac{1}{4} - 1 > 0 \implies (n - \frac{1}{2})^2 > \frac{5}{4}$$ Which is true, since $n \geq 4$ and therefore $(n - \frac{1}{2})^2 \geq \frac{49}{4}$.