Show that ${n}^{n^{n}}>n(n!)((n!)!)$ where n is a positive integer greater than or equal to $3$.

Question

My attempt:

Rewrite ${n}^{n^{n}} =n^{n} n^{n} … n^{n} n^{n} $ and $n(n!)((n!)!)=n(n)(n-1)(n-2)…1(n!)(n!-1)…1$.

Since $n^{n}\gt n!$, $n^{n} n^{n} … n^{n} n^{n}\gt n!n!…n!n!=n!^{n}$.

I tried to show that $n!^{n}\gt n(n)(n-1)(n-2)…1(n!)(n!-1)…1$, but I don’t know how.

Please give me a hint instead of a solution.

you typed in ${n^n}^{n}$ which is equal to $n^{n^2} $ I suspect you ought to have $n^{n^n} = n^{ \left(n^n \right)} $ So: what is the source of the problem? — Will Jagy, Feb 25 '24 at 02:52
Well we are a question-answer site, so this question will be receiving a full answer and not just a hint. — Mike, Feb 25 '24 at 02:52
@Mike Yes, I meant I prefer hint not that I you should not give an answer lol. — Chanhyuk Park, Feb 25 '24 at 03:03
Please write this correctly. You are meaning $(n^n)^n=n^{n^2}$? — Ted Shifrin, Feb 25 '24 at 03:37
Anyway the top part of my answer is the hint, the bottom part answers more fully. — Mike, Feb 25 '24 at 03:38
Since you typed {n^n}^{n}, I guess that what you mean is not n^{n^n} like in @Mike's answer. Please clarify your post by inserting parenthesis. — Anne Bauval, Feb 25 '24 at 03:38
I saw $n^{n^n}$ but the Mathjax was formatting funny. Sometimes it came out $n^{nn}$. — Mike, Feb 25 '24 at 03:39
Right. My bad. So please Chanhyuk Park edit your post to type correctly $n^{(n^n)}$. — Anne Bauval, Feb 25 '24 at 03:48
Chanhyuk Park: at the beginning of your attempt, what you are expanding is not $n^{(n^n)}$ as it should, but the (too small) $(n^n)^n=n^{nn}=n^{n^2}$. — Anne Bauval, Feb 25 '24 at 04:01
Why the hypothesis $n\ge3$? The strict inequality still holds for $n=2$. So, same as Will's very first comment: "what is the source of the problem"? — Anne Bauval, Feb 25 '24 at 04:16
@Mike In principle, yes it is! but I feel that parenthesis, though redundant, would be safer. — Anne Bauval, Feb 25 '24 at 22:34

Mike · Accepted Answer · 2024-02-25T03:45:28.087

HINT: Take logarithms to conclude $$n^{n^n} \ge ((n!)!)^4,$$ and note that this gives you what you need. En route, use the easy to see inequality $$M^M \ge 2^M (M!).$$

IF YOU NEED MORE ELABORATION:

First note that, taking logs of the LHS of the OP, we note: $$\log(n^{n^n}) = n^n \times \log (n).$$

On the other hand, taking logs of $(n!)!$ on the RHS of the OP:

$$\log((n!)!)$$ $$\le \log((n!)^{n!})$$ $$\le (n!)×\log(n!)$$ $$\le (n!)×\log(n^n)$$ $$\le (n!)\times (n \log n)$$ $$\le 2^{-n}n^n ×(n \log n),$$ Or in particular,

$$\log((n!)!) \le 2^{-n}n^n ×(n \log n). $$

However, for $n \ge 4$ note the inequality $$2^{-n} ×n \le \frac{1}{4}.$$ So plugging this into the above gives

$$\log((n!)!) \le \frac{n^n \log n}{4}.$$

However, we have already observed $$\log(n^{n^n}) = n^n\log(n),$$ and thus combining the above two relations yields $$\log(n^{n^n}) \ge 4(\log((n!)!), $$ or equivalently, $$n^{n^n} \ge (n!)!)^4.$$ Can you conclude yourself $$n × n! \le ((n!)!)^2$$ $$< ((n!)!)^3$$ yourself to get the desired result?

Show that ${n}^{n^{n}}>n(n!)((n!)!)$ where n is a positive integer greater than or equal to $3$.

1 Answers1