How to solve the inequality $n! \le n^{n-2}$? The inequality is $n! \le n^{n-2}$. I used Stirling's approximation for factorials and my answer was $n \le (e(2\pi)^{-1/2})^{2/5}$ but this doesn't seem right. Any help would be much appreciated.

Stirling says $n! \approx \sqrt{2\pi} n^{n-1/2} e^{-n}$.  Let $S(n)$ be the right side of that.  Now $S(n) \le n^{n-2}$ is equivalent to $\sqrt{2\pi} n^{5/2} \le e^n$, and that's certainly true for large enough $n$ (an exponential always beats a power in the end). To find exactly what $n$ makes the two equal requires using the Lambert W function: it turns out to be 
$$-\frac{5}{2}\,{\it W} \left( -1,-{\frac {{2}^{4/5}}{5\; {\pi^{1/5} 
}}} \right) \approx 4.883692535
$$
Of course this is only for the approximation, not the actual $n!$.  But in fact your inequality $n! \le n^{n-2}$ is true for $n \ge 5$ (note that $4! > 4^2$ and $5! < 5^3$).
