Find the convenient $n$ which make $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1$ be true. My question is: 

For which $n$ respect to $\alpha$ (where $\alpha\in\mathbb{R}$), the following inequality is true? 
  $$\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; ?$$
  or equivalently $\frac{n!}{n^{\alpha}}\geq \; \pi^n\;\;;\;n\geq\; ?$

To answer this question, at first I supposed that $\alpha\in\mathbb{N}$. So I arrive at the following
If $\alpha=1$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 9$;
If $\alpha=2$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 11$;
If $\alpha=3$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 14$;
If $\alpha=4$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 16$;
If $\alpha=5$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 18$;
If $\alpha=6$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 20$;
If $\alpha=7$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 21$;
If $\alpha=8$ then $\frac{n!}{n^{\alpha}\; \pi^n}\geq 1\;\;;\;n\geq\; 23$;
So I could not guess any closed form for $n$ which $\alpha\in\mathbb{N}$ case, and I think this method does not work even for integer case. Also I could not find any thing about $\alpha\in\mathbb{R}$.  Thanks for any useful help.
 A: May be, you could consider that you look for  $n$ such that
$$\frac{\log \left(\pi ^{-n}\, n!\right)}{\log (n)}=\alpha$$ and notice that, very quickly, the lhs is almost a straight line.
This seems to be interesting for Newton method. The problem being the starting guess, using your values for $\alpha=6$ and $\alpha=8$, let me propose the simple
$$n_0=17+\frac 32 \alpha$$
Trying for $\alpha=123$, the iterates would be 
$$\left(
\begin{array}{cc}
 k & n_k \\
 0 & 201.500 \\
 1 & 204.891 \\
 2 & 204.889
\end{array}
\right)$$
Still playing now for $\alpha=1234$, the iterates would be 
$$\left(
\begin{array}{cc}
 k & n_k \\
 0 & 1868.00 \\
 1 & 1731.55 \\
 2 & 1731.36
\end{array}
\right)$$
For sure, we could work a better guess $n_0$ but this seems to work.
Edit
Working backward, it seems that a better estimate could be
$$n_0=\frac{417}{98}+\frac{232 }{93}\alpha ^{39/80}+\frac{169 }{106}\alpha ^{39/40}$$
Using $\lceil n_0 \rceil$ for $\alpha=1,2,\cdots,8$ (cases you worked), this would give for $n_0$ the sequence $\{9,11,14,16,18,20,22,24\}$ which does not seem to be too bad.
Concerning the worked cases : for $\alpha=123$, this would give $\lceil n_0 \rceil=205$ and, for $\alpha=1234$, this would give $\lceil n_0 \rceil=1732$; these happen to be the solutions. 
For $\alpha=12345$, this would give $\lceil n_0 \rceil=15803$ while the exact solution would be $15862$.
Using this totally empirical formula for $n_0$, I suppose that one  Newton iteration would give the result.
Concerning the  derivative (required by Newton method), its formal expression is quite complex and I would suggest to compute it using finite differences that it to say to use
$$\frac d {dn} \left(\frac{\log \left(\pi ^{-n}\, n!\right)}{\log (n)}\right)=\frac 12\left(\frac{\log \left(\pi ^{-n-1}\, (n+1)!\right)}{\log (n+1)}-\frac{\log \left(\pi ^{-n+1}\, (n-1)!\right)}{\log (n-1)}\right)$$ or, even simpler, to approximate it using
$$\frac d {dn} \left(\frac{\log \left(\pi ^{-n}\, n!\right)}{\log (n)}\right) \approx 1-\frac{\log(\pi e)}{\log(n)}+\frac{\log(\pi e)}{\log^2(n)}$$
