$(n+1)!>n^2, \forall n\ge 4$

Question

The base case is clear, since if $n\gt 4, 5!=120\gt 25=5^2$

So assume $n=k$ which shows that $k!\gt k^2$.

Then if $n=k+1$,

$$(k+1)!=(k+1)k!$$ $$\gt (k+1)k^2 $$ Induction argument $$=k^3+k^2$$ $$\gt k^2+2k+1$$ $k^3>2k+1, \forall k\ge4$ $$=(k+1)^2$$

Is this a good proof? I never attempted the factorial induction proofs, and I first tried proving the other direction, that $n^2<n!$, and starting with $(k+1)^2$.

Answer 1

(Judging from your title, in which your result is $(n+1)!>n^2$) Your Induction Hypothesis should be $k!>(k-1)^2$, and thus you should be trying to prove $(k+1)!>k^2$, and you shouldn't bring in an inequality like $k^3>2k+1$ for $k\geq 4$ unless you've already proven it, so just make sure you prove that as a lemma somewhere in your proof.

But whatever the case, note that $(n+1)!=(n+1)n(n-1)!$. $(n+1)n=n^2+n>n^2$ for every $n>0$, so the result holds for more than just $n\geq 4$, it holds for $n\geq 1$.

Answer 2

$n=k\implies (k+1)!>k^2$

For $n=k+1, (k+1+1)!=(k+2)\cdot (k+1)!>(k+2)\cdot k^2$ which needs to be $\ge(k+1)^2$

$$\iff k^3+2k^2\ge k^2+2k+1\iff k^3+k^2-2k-1\ge0$$

Now, $\displaystyle k^3+k^2-2k-1\ge k^2-2k-3$ for $k>0$

which is $\displaystyle=(k+1)(k-3)>0$ for $k>3$

Answer 3

Edit:
Upun revisiting this answer I realise that it is not a proof by induction. It's a direct proof. It simply comes down to $$\left.\begin{matrix}(k+1)k>k^2\\(k-1)!>1\end{matrix}\right\}\implies\underbrace{(k+1)k\cdot(k-1)!}_{=(k+1)!}>k^2.$$ No induction needed or used.

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Original answer:

Suppose that for $n=k$ we have $(k+1)!>k^2$

for $n=k+1$ we have $(k+1+1)!=(k+2)!=(k+2)(k+1)k!$

since $(k+2)(k+1)>(k+1)^2$ obviously holds we also have $(k+2)(k+1)k!>(k+1)^2$ because $k!\geq1$, always. This proves that $(k+2)!>(k+1)^2$. Together with with the induction basis this proves your theorem.

Answer 4

$$(n+1)! \ge (n+1)n>n^2$$ So this is true for all $n\ge 0$