is my induction proof that $n!\gt n^2$ sufficient?

Question

question;

prove that $\forall\ n\ge4, n\in \mathbb{Z}, \ n!\gt n^2$.

my work;

let $n=4$ then $4!=24 \gt 4^2=16.$ true. now assume $n! \gt n^2$ is true for all $n\le k$

so now assume $k! \gt k^2$, so that the induction step is as follows,

$(k+1)!=(k+1)k! \gt k^2(k+1)=k^3+k^2 \gt k^2+2k+1=(k+1)^2$

which now shows that $n! \gt n^2$ holds for $(n+1)$ and is thus true for $n\ge 4$.

or do I have show that $k^3 \gt 2k+1$? which is clearly true $\forall k \gt 1, k \in \mathbb{Z}$.

Answer 1

To be 100% vigorous with the proof, you'd need to show that $k^3 > 2k +1$ for $k>4$. And since it's an easy one, go ahead and do it with induction again.

Otherwise, you're 100% spot on with your induction. Note that you don't need to use strong induction. Weak induction would suffice. It only needs to hold for $k$ into order to prove $k+1$. But that would not make your proof any more correct, just more concise.

Well done. :)