<s>[INST] Why do the rounded solutions of $n^x=x!$ follow this $2, 3, 3$ pattern as $n$ increases? Let $y$ be the solution to the following equation:
$n^x = x!$ where $n$ is an integer. I have observed that if you round $y$ to the nearest integer it follows a certain pattern. Let $c$ be the nearest integer to $y$. $c$ increases by $2, 3, 3$ as $n$ increases.
For example:
$$\begin{align}
1^x = x!, &\quad c = \phantom{1}1 \\
2^x = x!, &\quad c = \phantom{1}3 \quad(=\phantom{1}1+\color{red}{2}) \\
3^x = x!, &\quad c = \phantom{1}6 \quad(=\phantom{1}3+\color{red}{3}) \\
4^x = x!, &\quad c = \phantom{1}9 \quad(=\phantom{1}6+\color{red}{3}) \\
5^x = x!, &\quad c = 11 \quad(=\phantom{1}9+\color{red}{2})\\
6^x = x!, &\quad c = 14 \quad(=11+\color{red}{3})
\end{align}$$
Is this pattern true as $n$ goes to infinity? And, if so, why is this the case?
 [/INST] Your $n^x = x!$ is equivalent to $n = (x!)^{1/x}$
Stirling's approximation gives $x! \approx \sqrt{2\pi x}\left(\frac x e\right)^x$
so  $(x!)^{1/x} \approx (2\pi x)^{1/(2x)}\dfrac x e$
For large $x$ the first term heads to $1$ so for large $n$ you have $n \approx \dfrac x e$ and $x \approx n e$
Your $2,3,3$ pattern has an average of about $2.667$.  As $n\to \infty$ the average is in fact $e \approx 2.718$ so slightly more than two-thirds of the steps will be $3$
</s>