Which is greater, $98^{99} $ or $ 99^{98}$? Which is greater, $98^{99} $ or $ 99^{98}$?
What is the easiest method to do this which can be explained to someone in junior school i.e.  without using log tables.
I don't think there is an elementary way to do this. The best I could find was on Quora, in an answer by Michal Forišek on a similar question here, which is to consider $\frac{98^{99}}{99^{98}}=98.(\frac{98}{99})^{98}=98.(1-\frac{1}{99})^{98} \approx\ \frac{98}{e} >1$ and hence $98^{99} >  99^{98}$.
But the approx sign step does use definition of $e$ in terms of limits and thus cannot be considered elementary. Any other way?

Edit- I was hoping something that does not involve calculus, that is why I tagged it in number theory, but as it seems it is almost impossible to avoid calculus when exponentials are involved.
All the answers are fine, and can be explained to students in classes above seventh or eigth. My aim for this qustion was to check with you all, if I have missed some elementary trick or not,I guess I did not. I was looking for the easiest solution someone can come up with. Thanks!
 A: $$\frac{98^{99}}{99^{98}}=\frac{98}{\left(1+\frac{1}{98}\right)^{98}}>\frac{98}{97}>1$$
A: We could try and generalize a little bit and ask ourselves if $n ^ {n+1}$ is always bigger than $(n+1) ^ n$ where n is a natural number.
This turns into verifying if the inequality $n^{n+1} > (n+1)^n$ holds.
The left hand side can be rewritten as $n \cdot n^n$ and the inequality becomes $n \cdot n^n > (n+1) ^ n$
Being $n^n$ always positive, the inequality holds dividing both sides by it so we have $n > (\frac{n+1}{n})^n$ which can be written also as $n > (1+\frac{1}{n})^n$.
The term $(1+\frac{1}{n})^n$ tends to $e$ when n tends to infinity, therefore if $n \geq 3$, the latter inequality holds as the right hand side cannot be more than 2.71...
It does not hold for n < 3, in fact 1^2 (=1) < 2^1 (=2) and 2^3 (=8) < 3^2 (=9). For 3 we have 3^4 (=81) > 4^3 (=64) and so on.
A: Using Bernoulli's inequality:
$$ (1+x)^n\ge 1+nx \qquad \forall \ \text{$x\ge -1$ and $n\in\mathbb N_0$ }$$
we have
$$ \left(1-\frac1{99}\right)^{49}\ge 1-\frac{49}{99}=\frac{50}{99}>\frac{50}{100}=\frac12$$
Therefore
$$ \frac{98^{99}}{99^{98}}=98\cdot \left(1-\frac1{99}\right)^{49}\cdot \left(1-\frac1{99}\right)^{49}>98\cdot \frac12\cdot\frac12>1.$$
A: Hint We can write
$$\frac{99^{98}}{98^{99}} = \frac{1}{98}\left(1 + \frac{1}{98}\right)^{98}.$$
Now, one can show via an easy induction (if you find yourself stuck, see this one-line proof) that $\left(1 + \frac{1}{n}\right)^n < n$ for $n \geq 3$, so the above quantity is less than $$\frac{1}{98} (98) = 1 ,$$ and rearranging gives
$$99^{98} < 98^{99}.$$
A: I hope at least we can use the Binomial Theorem with integer exponents.
I will also assume knowledge of the usual $\binom mn$ notation for the coefficients, but that is not really necessary; you can rewrite the
argument to do without that notation.
Observe that for $m \geq 1$,
$$\binom{98}{m + 1}
 = \frac{98 \cdot 97 \cdot 96 \cdot \cdots \cdot (98 - m)}
        {1 \cdot 2 \cdot 3 \cdot \cdots \cdot (m + 1)}
 = \frac{98 - m}{m+1} \binom{98}{m} \leq \frac{98}{2} \binom{98}{m}.
$$
from which we can conclude that for $m \geq 2$,
$$\binom{98}{m} \leq \left(\frac{98}{2}\right)^{m-1} \binom{98}{1}
\leq \frac{1}{2^{m-1}}\cdot 98^m \leq \frac12\cdot 98^m.$$
Therefore
\begin{align}
99^{98} &= (98+1)^{98} \\
&= 98^{98} + \binom{98}{1} 98^{97}
     + \underbrace{ \binom{98}{2} 98^{96}
        + \binom{98}{3} 98^{95} +  \cdots
        + \binom{98}{97} 98 + 1}_{97 \text{ terms}} \\
& \leq 98^{98} + 98^{98}
      + \underbrace{ \frac12\cdot 98^{98} + \frac12\cdot 98^{98} + \cdots
           + \frac12\cdot 98^{98} + \frac12\cdot 98^{98}}_{97 \text{ terms}} \\
& \leq \left(2 + 97 \cdot \frac12\right)98^{98} \\
& < 98 \cdot 98^{98} = 98^{99}.
\end{align}
So $98^{99}$ is greater.
A: $98^{99}>99^{98}\Leftrightarrow 98^{\frac {1}{98}}>99^{\frac {1}{99}}$ and $(n^{\frac {1}{n}})$ is strictly decreasing.
A: The inequality $99^{98} < 98^{99}$ is equivalent to
$$\bigl( 1 + \frac{1}{98} \bigr)^{98} < 98
$$
I'll prove by induction that
$$\bigl( 1 + \frac{1}{n} \bigr)^n < n \quad\text{if $n \ge 3$}\quad\text{(it's false if $n=1$ or $2$).}
$$
The basis step $n=3$ is $\frac{64}{27} < 3$.
So let's assume that 
$$\bigl( 1 + \frac{1}{n} \bigr)^n < n \quad\text{and}\quad \bigl( 1 + \frac{1}{n+1} \bigr)^{n+1} \ge n+1
$$
and derive a contradiction. These two inequalities can be rewritten 
$$\left( \frac{n+1}{n} \right)^n < n \quad\text{and}\quad \bigl(\frac{n+1}{n+2}\bigr)^{n+1} \le \frac{1}{n+1}
$$
Multiplying them we get
$$\left( \frac{n^2 + 2n + 1}{n^2 + 2n} \right)^{n} \cdot \frac{n+1}{n+2} < \frac{n}{n+1}
$$
$$\left( \frac{n^2 + 2n + 1}{n^2 + 2n} \right)^{n} \cdot \frac{n^2+2n+1}{n^2+2n} < 1
$$
$$\left( \frac{n^2 + 2n + 1}{n^2 + 2n} \right)^{n+1} < 1
$$
which is absurd.
A: The only "elementary" way I can think of is to write $99^{98} = (98 + 1)^{98}$ and then expand using the binomial expansion formula, and then show you get a sum of $99$ terms where each term is less than or equal to $98^{98}$, and the sum of the last two terms $98 + 1$ is strictly less than $98^{98}$. Then your sum is strictly less than $98 \cdot 98^{98} = 98^{99}$.
A: Let $$\displaystyle f(x) = x^{\frac{1}{x}}\;,$$ where $x>0$
Now $$\displaystyle f'(x) = x^{\frac{1}{x}}\cdot \left[\frac{1-\ln (x)}{x^2}\right]$$
So here $f'(x)>0$ for $x<e$ and $f'(x)<0$ for $x>e\approx 2.714$
So function $f(x)$ is an Strictly Decreasing function for $x>e\approx 2.714$
So $$f(98)>f(99)\Rightarrow 98^{\frac{1}{98}}>99^{\frac{1}{99}}\Rightarrow 98^{99}>99^{98}$$
A: you should know that the maxima of the function $$f(x)=x^\frac 1x$$ attains maxima at $x=e^\frac 1e$ 
So, as 98 is closer to $e$, hence we have $98^\frac 1{98}\gt 99^\frac 1{99}$
So, $98^{99}\gt99^{98}$
A: This is the most elementary proof I can muster, and assumes students are familiar with inequalities and the interest formula.
Assume $ 98^{99} > 99^{98} $ .
The following algebra holds:
$$ 98^{99} > 99^{98} $$
$$ 98 * 98^{98} > 99^{98} $$
$$ 98 > (\frac{99}{98})^{98} $$
$$ 98 > (1 + \frac{1}{98})^{98} $$
You could ask your students to compute $ (1 + \frac{1}{98}) ^{98} $ on their calculators, which will do so because the number is not excessively large. Or you could explain that this mysterious, 2.71 number actually appears with every $ (1 + \frac{1}{n})^{n} $, n greater than a certain value. Either way the assumed statement is proven true with the calculation.
In either case, the formula $ (1 + \frac{1}{n})^{n} $ is elementary since it computes basic interest over time. I learned this in grade school around the age of 16 I would say...
A: I have generalized a bit more this result finding when $$n^{n+m} > (n+m)^n$$
Of course $n^{n+m} = n^n \cdot n^m$ and being $n^m > 0$ it is possible to divide both sides of the inequality to find that $$n^m > (\frac{n+m}{n})^n$$ that becomes $$n^m > (1+\frac{m}{n})^n$$
Now $lim_{n \to \infty} (1+\frac{m}{n})^n  = e^m$ (https://en.wikipedia.org/wiki/List_of_limits) so eventually we have $$n^m > e^m$$ therefore for $m>1$ the same condition as before holds: from $n>=3$ the condition is verified.
For instance, suppose $n=3$ and $m=2$ : we can state that $3^{3+2} > (2+3)^3 \rightarrow 3^5 > 5^3 \rightarrow 243 > 125$.
A: @ Bhaskar Vashishth: If you want a best answer as you comment to Lee Mosher take a graphic of $f(x)= x^{\frac{1}{x}}-1-\frac1{x}$and you can see that your inequality can be generalized to all real greater than $2.293$ for which this function is positive. Obviously, after to see that $(x+1)^x<x^{x+1}$ is easily equivalent to $x^{\frac{1}{x}}>1+\frac1{x}$. I feel all generalization of your inequality has been started from this graphic or some other equivalent.On the other hand, thinking a little about it seems to me that this should be something obvious for everybody.

