Prove $n!>a^n$ for all $a \in \mathbb{Z}^{+}$ with large enough $n$

Question

I want to prove that for all $a \in \mathbb{N}$, $a^n < n!$ for all large enough $n$.

I want to base my ideas off this similarly stated question: How to prove $n! > n^a$ for all $a\in \mathbb{R}$ (for sufficiently large $n$)? (but for me, the exponential has a swapped base and exponent)

Looking at this question, it seems that I should have started with $5^n < 5!$: How to prove $a^n < n!$ for all $n$ sufficiently large, and $n! \leq n^n$ for all $n$, by induction? (But I think mine is for any $a$ rather than a specifc one?)

• This answer seems almost exactly what I was going for, but I'm not sure how the function for $n_0$ was determined: https://math.stackexchange.com/a/1769051/454931

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For my question, I first started by looking at some small cases, when $a=2$, then $n=4$, and then $a=3$, then $n=7$. (and $4$ with $9$, then $5$ with $12$, etc.)

So I think if I let $n \geq 2a$ then this statement should hold?

I'm not sure how to set up the induction though. Would I start with $a^n < n!$ and maybe use $a=2$ as my base case and have the induction hypothesis hold because $a^k < k!$ for all $k \geq 2a$

Answer 1

The statement will not necessarily hold when $n>2a,$ for example, for $a=5,$ $n=11$ doesn't work (but $n=12$ does).

I am not sure whether $\exists\ x\in\mathbb{R}$ such that, $\forall\ a\in\mathbb{N},\quad n>xa\ \implies a^n<n!.$

But what is easy to see and along the lines of what you want, I think, is the following:

Proposition: $\ \forall\ a\in\mathbb{N},\ n>a^2\implies a^n<n!. $

We first prove a Lemma, which we will use in the proof.

Lemma: If $a,b\in\mathbb{Z}$ and $2\leq b\leq a^2-1,$ then $b(a^2+1-b)> a^2.$ Proof: $a^2b-a^2 = a^2(b-1)\geq(b+1)(b-1) = b^2-1 > b^2 - b.$ Rearranging gives: $a^2b + b - b^2 > a^2.$

Proof of proposition: By the Lemma, for even $a,\quad a^{(a^2)} = (a^2)^{a^2/2} < (1\times a^2)(2\times (a^2-1))\cdots \left(\frac{a^2}{2} \times (\frac{a^2}{2}+1) \right) = (a^2)! $ We get something similar for odd $a,$ although details are left to the reader/ the less lazy than I. Therefore, for $n>a^2,\quad a^n = a^{a^2+k} = a^{a^2} a^k$ for some positive integer $k.$ But $a^{a^2} a^k < (a^2)!a^k < (a^2)!(a^2+1)(a^2+2)\cdots (a^2+k) = (a^n)!,$ and so the result follows.

Answer 2

From Stirling's formula $$ \frac{n!}{a^n}=\frac{n^n e^{-n}\sqrt{2\pi n}}{a^n}(1+o(1)) =\left(\frac{n}{ae}\right)^n\sqrt{2\pi n}(1+o(1))\to\infty $$ as for $n\ge ae$ this ratio is at least $\sqrt{2\pi n}$.

Answer 3

Let $m=\lceil a\rceil$. $$\frac{1\cdot2\cdot3\cdot4\cdots n}{a\cdot a\cdot a\cdot a\cdots a}=\frac{1\cdot2\cdot3\cdot4\cdots m\cdots n}{a\cdot a\cdot a\cdot a\cdots a\cdots a}>\frac{1\cdot2\cdot3\cdot4\cdots m}{a\cdot a\cdot a\cdot a\cdots a}\left(\frac{m+1}a\right)^{n-m}.$$

As $\dfrac{m+1}a>1$, there will be a power of this ratio that makes the RHS larger than one, and from then on, the ratio keeps increasing.

E.g., $a=\pi\to m=4$. Then $\dfrac{4!}{\pi^4}=0.246\cdots$ and $\dfrac 5\pi=1.591\cdots$. We have $\dfrac{4!}{\pi^4}\left(\dfrac 5\pi\right)^4>1$ so that for all $n\ge4+4$, $n!>\pi^n$.