I proved this in the special case $x = 99, y = 100$, here. As others have pointed out, what you really want to hold is the following:
Statement: Let $x, y \in \mathbb{R}$. Then $y > x > e$ implies $x^y > y^x$.
Proof:. Write $y = x + z$, where $z > 0$. Then,
$$\begin{align}
x^y > y^x &\iff x^x x^z > y^x
\\
&\iff x^z > \left(\frac{x+z}{x} \right)^x
\\
&\iff x^z > \left( 1 + \frac{z}{x} \right)^x.
\end{align}$$
The right hand side $\left(1 + \frac{z}{x} \right)^x$ is monotone increasing with limit $e^z$. Since the left hand size is strictly greater than $e^z$ (as $x > e$), it follows that the inequality always holds.