Prove by induction. Show that if $n ≥ 6$, then $n! > n^3$ Show that if $n ≥ 6$, then $n! > n^3$
Initial Step: $n = 6$
LHS: $6!=720$
RHS: $6^3=216$
LHS > RHS
Inductive Step: Assume $n=k$ is true
$k! > k^3$
Prove $n=k+1$ is true
$(k+1)! > (k+1)^3$
Can you help me? I don't know where to go from here. I'm stuck in here.
 A: Assume that $k! > k^3$ ...$(1)$
Let's show $(k+1)k! > (k+1)^3$
Multiply $(1)$ by $(k+1)$
$(k+1)k! > (k+1)k^3$
$(k+1)! > (k+1)k^3>(k+1)(k+1)^2=(k+1)^3$
Using
$k^3>(k+1)^2$
$k^3-k^2+2k-1>0$, $k^3>k^2$, and,  $2k-1>0$ given that $k\geq6$
A: It suffices to show that $k^3>(k+1)^2$ for $k≥6$ and you can do that using induction.
A: Base case of $k=6$ proved.
Moving forward, LHS increases by a factor of $(k+1).$
RHS increases by a factor of $\left(\frac{k+1}{k}\right)^3.$
Although it can be argued that $k=7$ and $k=8$ must be independently verified,
thereafter, with $\frac{k+1}{k} < 2$,
RHS increasing by smaller factor than LHS.
A: \begin{align}
(k+1!) &> (k+1)^3 \\
k! (k+1) &> (k+1)^3 \\
k! &> (k+1)^2
\end{align}
Our inductive hypotesis is $k! > k^3$.
$$ k^3 > (k+1)^2 $$
$$ k^3 - (k+1)^2 > 0 $$ (1)
$2.148$ is a root of $ f(k) = k^3-(k+1)^2$.
Dividing $f(k)$ by $(x - 2.148)$ you can see that the discriminant of the second order polinomial is negative, so it's the only root. Also, it's derivarive $f'(k) = 3k^2 - 2(k+1)$ is positive for $k> 1.215$.
So, (1) is true for $k> 2.148$. In particular, for $k>6$.
