Proof by induction: $n! > n^2$ for $n\ge 4$

Question

Proof by induction: $n! > n^2$ for $n\ge 4$

Basis step: if $n=4$

$4\cdot3\cdot2\cdot1 > 4^2$

$24 > 16$

I don't know how to do the inductive step.

Answer 1

The inductive step:

$$(n+1)!=(n+1)n!\ge(n+1)n^2\ge (n+1)^2$$ and to justify the last inequality notice that the quadratic polynomial $x^2-x-1\ge0\;\forall x\ge4$ since its discriminant $\Delta=5$ and the roots are $x_1=\frac12(1-\sqrt5)<0$ and $x_2=\frac12(1+\sqrt5)<4$ so the polynomial is positive for all $x$ greater than $x_2$.

Answer 2

Inductive step:

$$(n+1)! = (n+1)n! > (n+1)n^2 $$

But clearly, $n^2 > n + 1$ for integer $n \ge 2$. You could prove this rigorously by showing that the curve $n^2 - n - 1$ lies strictly above the $x$ axis for $n \ge 2$.

Hence,

$$(n+1)! > (n+1)^2$$

Answer 3

Inductive step

Assume it holds for $k$

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