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I am trying to prove that:

$n^n \geq n!$ is valid for a $x$ set of numbers.

So, I am trying an inductive process.

However, case $P(0)$ doesn't seem to work because I have read somewhere that $0^0$ is undefined, but I've also read that $0^0$ is $1$.

What approach do you recommend?

Martin Sleziak
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JOX
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  • I do not know what is there to prove. For any positive integer n,

    n! = n(n-1)(n-2)... 1 (n terms)

    n^n = nnn*..........n (n terms).

    n > n-1, n > n-2, ... n > 1

    Thus for any positive n, n^n >= n!.

    Did you mean to prove that if n! is divisible by m! then, n >=m?

    Thanks

    Satish

     – Satish Ramanathan Nov 26 '13 at 04:02
  • @satishramanathan You just provided a proof of what needed to be proved (though perhaps the OP should provide a little bit more elaboration if he were to use this proof). – John Patterson Nov 26 '13 at 04:05
  • If the convention you adopt leaves $0^0$ undefined, then $n^n\ge n!$ doesn't make sense when $n=0$ in the first place. Start your induction at $n=1$ instead. –  Nov 26 '13 at 04:17
  • The problem is that using cases 0 and 1 as 'bases' gives me the following conclusion: n^n = n!, how do I determine the case can also be >? – JOX Nov 26 '13 at 04:52
  • See https://math.stackexchange.com/questions/1260233/inductive-proof-that-kkk-for-k-geq-2 – Arnaud D. Nov 29 '19 at 14:12

3 Answers3

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$0^0$ is most definitely not undefined and $0^0 = 1$ as you've read. If you want a more thorough explanation this seems like a nice article.

EDIT: As pointed out in the comments by T. Bongers, my first sentence is misleading. Outside of the world of analysis, you are fairly safe in holding with the definition. As highlighted in the article there are very good reasons by we define $0^0 = 1$ in our usual context (here we are actually considering $f(x)^{g(x)}$ as $\lim_{x \to 0^+}{f(x)}$ and $\lim_{x \to 0^+}{g(x)}$). It is perhaps, a simplification, but a thoroughly discussed and justified simplification for our everyday application.

Induction is a simple method of proof:

Basis: $P(0) = 0^0 = 1 \ge 0!$.

Assume $P(k)$ holds or that $k^k \ge k!$. So, for $k+1$:

We know $(k+1)^{(k+1)} = (k+1)^k(k+1)$ and that $(k+1)! = k!(k+1)$. So, $(k+1)^k \ge k^k \ge k!$ and so we multiply through by $(k+1)$, so $(k+1)^k(k+1) \ge k^k(k+1) \ge k!(k+1)$ hence $(k+1)^k(k+1) \ge k!(k+1)$. Using our earlier equalities, however, this is the same as $(k+1)^{(k+1)} \ge (k+1)!$. $\Box$

John Patterson
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    I think your beginning sentence is misleading: The symbol $0^0$ may be assigned the value $1$ for special contexts such as this, but $0^0$ is not defined in general. –  Nov 26 '13 at 04:18
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    @T.Bongers I have added a further explanation following the first paragraph.Perhaps this alleviates some of the subterfuge of my simple first sentence. I didn't want to confuse him further than I perhaps was about to, but thoroughness is good. – John Patterson Nov 26 '13 at 04:43
  • can I assume P(0)1≥1 even if it is 1=1? – JOX Nov 26 '13 at 06:20
  • @DanielOrtizCosta I'm not sure if I'm interpreting your question correctly, so please adjust my understanding if it is innacurate: you are asking does $1 \ge 1$ hold as $1 = 1$? The answer is yes. Because of the bar under the greater than symbol, the statement $x \ge y$ is true if and only if $x \gt y \lor x=y$. Notice this is a logical or, so you must only show that $x=y$ or $x>y$ follows from the premise, you do not have to show both (that would be a paradox, actually). – John Patterson Nov 26 '13 at 14:02
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Full formulation of the proof alluded to in Satish's hint:

Proving that $n^n \geq n!$ is the same as proving that $\frac{n^n}{n!} \geq 1$ for all $n$. For $n > 0$, note that $$\frac{n^n}{n!} = \frac{n}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{n}{2} \cdot \frac{n}{1}$$

Note that each term on the right side is greater than or equal to $1$, as the denominator is always less than or equal to the numerator. A product of values greater than or equal to $1$ is surely greater than or equal to $1$, giving us the result for all values of $n$ greater than $0$. For $0$, as stated above, the equality holds.

Cheers!

HJ32
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If $a > b$, then $ac > bc$. This is all you need to set up the induction proof. Start with $n=1$. If you need to extend to $n=0$, then you can choose the definition $0^0 = 1$ if you see fit.

Emily
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