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

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