Which is larger, $99^{100}$ or $100^{99}$?

Which is larger, $999^{1000}$ or $1000^{999}$?

Or more generally, for $n$ any natural number: which is larger, $n^{n+1}$ or $(n+1)^n$?

Please do not use a computer, calculator, or any electronic aid to work out the answer for any specific cases of $n$. You should be able to solve this puzzle purely by mathematical reasoning and without calculations, using a computer only to read the question and type the answer :-)

  • 7
    $\begingroup$ I think the math methods needed to solve this are too standard for this to be an interesting puzzle. $\endgroup$
    – xnor
    Jul 8, 2015 at 1:03
  • 1
    $\begingroup$ I respectfully disagree with xnor -- I thought the connection needed to solve the puzzle was interesting and unexpected. Thanks for posting! $\endgroup$ Jul 8, 2015 at 5:38
  • $\begingroup$ It's easy for people like me and @xnor who have lots of advanced maths knowledge, but interesting for those who don't. Maybe xnor is just sore he didn't get to this one first? :-) $\endgroup$ Jul 8, 2015 at 8:42
  • $\begingroup$ How are we to compare their pixel size to see which is larger without the use of computers?! You are a madman! $\endgroup$
    – Mark N
    Jul 8, 2015 at 12:57

4 Answers 4


$n^{(n+1)}$ is larger iff $n\geq 3$

$$n^{(n+1)} > (n+1)^n \\n > \left(\frac{n+1}{n}\right)^n \\n > \left(1+\frac 1 n\right)^n$$

The right side increases with $n$, and we know that

$$\lim_{n\to\infty}\left(1+\frac 1 n\right)^n=e$$

Therefore, for $n\geq 3$,

$$n > e > \left(1+\frac 1 n\right)^n$$

Now we only have to check $n=2$ and $n=1$ by hand.

$$2^3 < 3^2\\1^2 < 2^1$$

  • $\begingroup$ Very good. This is what I was looking for! $\endgroup$ Jul 8, 2015 at 8:43
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    $\begingroup$ @2012rcampion - could you explain to me how you get from the first line to the second in your work? $\endgroup$
    – Nyk 232
    Jul 8, 2015 at 13:58
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    $\begingroup$ @Nyk232 You divide by $n^n$ on both sides. $\endgroup$
    – mmking
    Jul 8, 2015 at 14:14
  • $\begingroup$ @mmking ahh, I see now. thanks for letting me know! $\endgroup$
    – Nyk 232
    Jul 8, 2015 at 14:15

Asking which of $x^y$ or $y^x$ is bigger is the same as asking which of their logs is bigger, so what comparison sign should fill the blank in $$ y\log x\,\,\,\_\,\,\,x\log y? $$ Rearranging, this becomes $$ \frac{\log x}{x}\,\,\,\_\,\,\,\frac{\log y}y $$ So, letting $f(t)=\frac{\log t}t$, is $f(x)$ or $f(y)$ bigger? Looking at the graph of f(t), we see that $f$ is increases up to a point, then decreases thereafter. To find this point, we compute the derivative $f'(t)=\frac{1-\log t}{t^2}$, then set this to zero, so the maximum of $f$ occurs at $t=e$. In summary:

$x^y-y^x$ has the same sign as $f(x)-f(y)$, which is negative when $x<y<e$, and positve when $e<x<y$.

In our case, we see that $n^{n+1}-(n+1)^n$ will be positive when $n\ge3$, so for almost all $n$, $n^{n+1}>(n+1)^n$.

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    $\begingroup$ I think "looking at the graph" counts as using a computer... but it's very easy to show that $f(t)$ is only increasing when $$f'(t)>0\\\frac{1-\log t}{t^2}>0\\1-\log t>0\\\log t<1\\t<e$$ (for positive $t$). $\endgroup$ Jul 8, 2015 at 3:12
  • $\begingroup$ @2012rcampion very good points! $\endgroup$ Jul 8, 2015 at 3:14
  • $\begingroup$ Another interesting approach, +1 $\endgroup$ Jul 8, 2015 at 8:43
  • $\begingroup$ @2012rcampion Graph scetching does not need a computer. Also you should not really need to look at the graph of log(t)/t to see that it will increase up to a point maximum and then turn. $\endgroup$
    – Taemyr
    Jul 8, 2015 at 11:22

The question said no calculations as well as no calculators, so here is my answer with the fewest calculations I could manage.

Ask another question: Which pair is going to be closer,

$99^{100}$ and $100^{99}$


$999,999^{1,000,000}$ and $1,000,000^{999,999}$?

99 is almost 100 only a bit smaller, so doing another multiplication is making a number that is a bit smaller an other 100 times larger.

999,999 is much closer to a million that 99 is to 100 so the effect of raising an extra power will be much closer to multiplying by an extra million adding 6 zeros.

Clearly $n$ being larger favours $n^{n+1}$ to be the greater of the two.

So how small does $n$ have to be for $n^{n+1}$ to be smaller.

$1^{2} - 2^{1} = 1 - 2 = -1$
$2^{3} - 3^{2} = 8 - 9 = -1$
$3^{4} - 4^{3} = 81 - 64 = 17$

Looks like $n^{n+1}$ is biggest when...

n > 2

Negative values of $n$ exhibit the switching of sign expected for multiplying negative numbers, so there is a different rule, but that is left as an exercise, as the question only asked for natural numbers (positive whole numbers).

  • $\begingroup$ Even 81-64 is a bigger calculation than anyone else had to do. I don't agree that this involves less calculation than the other answers. $\endgroup$ Jul 8, 2015 at 8:44
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    $\begingroup$ @randal'thor But you don't need to calculate that, you only need to see that 81 is larger than 64, and that's trivial. $\endgroup$
    – Dorus
    Jul 8, 2015 at 9:07
  • $\begingroup$ @Dorus Well, even calculating 4^3 and 3^4 then! $\endgroup$ Jul 8, 2015 at 9:10
  • $\begingroup$ @Bob It's a standard precalc identity: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty}\exp\left\{n{}{\log}\left(1+\frac{1}{n}\right)\right\}$$ $$=\exp\left\{\lim_{n\to\infty}n\log\left(1+\frac{1}{n}\right)\right\}$$ $${}{}=\exp\left\{\lim_{n\to\infty}\frac{\log\left(1+\frac{1}{n}\right)}{n^{-1}}\right\}$$ $$=\exp\left\{\lim_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^{-1}(-n^{-2})}{-n^{-2}}\right\}$$ $$=\exp\left\{\lim_{n\to\infty}\frac{n}{n+1}\right\}$$ $$=\exp(1)=e$$ $\endgroup$ Jul 8, 2015 at 13:30

For sufficiently large $n$, we can check via division and limits.

$$ \frac{n^{(n+1)}}{(n+1)^n} \\ =\frac{n^{(n+1)}}{(n^n+j*n^{n-1}+k*n^{n-2}+...)} \\ =\frac{n^{(n+1)}}{(n^n+j*n^{n-1}+k*n^{n-2}+...)}*\frac{1/n^n}{1/n^n} \\ =\frac{n}{1+j/n+k/n^2+...} \\ \lim_{n\to\infty}\frac{n}{1+j/n+k/n^2+...}=\infty $$

So we can conclude that for sufficiently large $n$, $$n^{(n+1)} > (n+1)^n \\$$

2012rcampion did the work to show the threshold, where $n$ must be greater than 2.

  • 2
    $\begingroup$ This proof doesn't work. The value $j$ and $k$ depend on $n$, and the number of summands also grows with $n$. $\endgroup$
    – xnor
    Jul 8, 2015 at 0:57

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