Which is greater, $98^{99} $ or $ 99^{98}$? [duplicate]

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!

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 $$ \frac{98^{99}}{99^{98}}=98\cdot \left(1-\frac1{99}\right)^{49}\cdot \left(1-\frac1{99}\right)^{49}>98\cdot \frac12\cdot\frac12>1.$$

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.

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 $98 \cdot 98^{98} = 98^{99}$.

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 $$f(98)>f(99)\Rightarrow 98^{\frac{1}{98}}>99^{\frac{1}{99}}\Rightarrow 98^{99}>99^{98}$$