I was looking at the ratio $$\frac{k!}{k^k} =\frac{1*2*3*...*k}{k*k*k*...*k}<1$$ and I wondered how large of a factorial I would need to make the ratio greater than one. So, $$\frac{n!}{k^k} =\frac{1*2*3*...*k*...*n}{k*k*k*...*k}>1$$ I have found that since $$\frac{(2k)!}{k^k}=\frac{1*2*3*...*k*...*2k}{k*k*k*...*k}$$the right half of the factors in the numerator are all greater than $k$, and we have that ${(2k)!}>{k^k}$.
Therefore, n is bounded by $$k<n<2k.$$ I have attempted to improve these bounds but have not been able to make progress.$$$$ I started with the inequality:$$k^k<n!$$ and taking the log of both sides gives $$k*log(k)<log(n!)$$ From here, Stirling's Approximation takes away the factorial and replaces it with $$k*log(k)<n*log(n)-n+O(log(n))$$ After playing around with this last inequality for a while, I still have not been able to make improvements on the original bounds. How could I go about doing this? Thank you.
We have $$k \log (k) = n (\log (n)-1) + \frac 12\log(2\pi n)\\ n=\frac {k \log(k)- \frac 12\log(2\pi n)}{\log(n)-1}$$ We can bound $n$ by removing the negative term in the numerator and reducing the denominator, so $$n\lt \frac {k\log(k)}{\log(k)-1}$$ This appears to be reasonably close as $k$ gets larger. For $k=20$ it gives $n \approx 30$ where $26! \gt 20^{20}$ and for $k=50$ it gives $n \approx 67$ where $62! \gt 50^{50}$ You can get a better result by iterating to convergence. Let $n_0$ be the result from above, then iterate $$n_{i+1}=\frac{k\log(k)-\frac 12 \log(2 \pi n_i)}{\log(n_i)-1}$$ and round up. It looks like it works starting at $k=7$, though convergence is slow there. Convergence speeds up as $k$ gets larger.