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.

  • $\begingroup$ Can you prove $(7/6)^3<2\le k+1$ for $k\ge6$? What does that tell you about $((k+1)/k)^3$? $\endgroup$ – J.G. Oct 6 '20 at 16:53

Assume that $k! > k^3$ ...$(1)$

Let's show $(k+1)k! > (k+1)^3$

Multiply $(1)$ by $(k+1)$

$(k+1)k! > (k+1)k^3$

$(k+1)! > (k+1)k^3>(k+1)(k+1)^2=(k+1)^3$



$k^3-k^2+2k-1>0$, $k^3>k^2$, and, $2k-1>0$ given that $k\geq6$


It suffices to show that $k^3>(k+1)^2$ for $k≥6$ and you can do that using induction.


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.


\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 $k>6$.


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