Prove that $n$ factorial is larger than $n$ cube if $n$ is large enough

Question

I try to solve it by Mathematical Induction. However I don't know how to prove that $(k+1)!>(k+1)^3$

Answer 1

The precise statement is

If $n \ge 6$, then $n! > n^3$.

The induction step goes as follows:

$$(n+1)! = (n+1) n! > (n+1) n^3$$

and so it suffices to prove that $(n+1) n^3 > (n+1)^3$, or equivalently, $(n - 2) n (n + 1)>1$, which is clearly true if $n \ge 3$.

Answer 2

Consider the ``function'' $n!$ for $n\geq 5$, then we know that $$n!-n^3\geq n(n-1)\dots(n-4)+n^3=n^5+\dots$$ which is a polynomial function of order $5$, (which is odd).

We know that odd polynomials diverge as $n\to \infty$ thus the function will in particular be positive for $n$ large enough.

Comment: This proof applies to any power $n^k$.

Answer 3

Choose $k$ large enough that $3k^2+3k+1<k^3\implies (k+1)^3<2k^3<k!2<(k+1)!$. Since $k^3<k!$ has minimum solution $0$ (among non-negative $k$) and next-to-minimum solution $6$, you need only note $k\ge 6\implies 3k^2+3k+1<3k^2+4k<4k^2<k^3$.