How to solve the following limit similar to factorial type Let $\alpha>1$ be a fixed real number and $M(x)=\max\left\{m\in\mathbb{N}:m!\le\alpha^x\right\}$, prove that$$\displaystyle\lim_{n\to\infty}\displaystyle\frac{\sqrt[n]{M(1)M(2)\cdots M(n)}}{M(n)}=e^{-1}.$$ Use
$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n.$$ I got that $$\lim\limits_{n\to\infty}\dfrac{\sqrt[n]{n!}}{n}=e^{-1}.$$ But I don't know how to go on. The properties of this function $M(x)$ should be critical.
Is there any way to parse this function?
 A: Denote $A_n:=\displaystyle\frac1n\sum_{k=1}^n\log M(k)-\log M(n)$, and let $K_m:=\displaystyle\left\lceil\frac{\log m!}{\log\alpha}\right\rceil$.
Then $M(k)=m$ if and only if $K_m\leqslant k<K_{m+1}$. Further (summing by parts), $$\sum_{k=1}^{K_m-1}\log M(k)=\sum_{n=1}^{m-1}(K_{n+1}-K_n)\log n=K_m\log m-S_m,\\S_m:=\sum_{n=2}^m K_n\log\frac{n}{n-1}.$$
Therefore, for $K_m\leqslant n<K_{m+1}$, we have the estimates $$\frac{K_m\log m-S_m}{K_{m+1}}-\log(m+1)\leqslant A_n\leqslant\frac{K_{m+1}\log(m+1)-S_{m+1}}{K_m}-\log m.$$
Now, using Stirling's asymptotics for $m!$ (or even weaker estimates), we prove $$\lim_{m\to\infty}\left(\frac{K_m}{K_{m+1}}\log m-\log(m+1)\right)=0,\\\lim_{m\to\infty}\left(\frac{K_{m+1}}{K_m}\log(m+1)-\log m\right)=0,\\\left[\lim_{m\to\infty}\frac{S_m}{K_{m+1}}=\lim_{m\to\infty}\frac{S_{m+1}}{K_m}={}\right]\lim_{m\to\infty}\frac{S_m}{K_m}=1$$ (together, these give the needed $\color{blue}{A_n\to-1}$ as $n\to\infty$).
Say, for the last one, we use $$K_n\log\frac{n}{n-1}=\left(\frac{n\log n}{\log\alpha}+O(n)\right)\left(\frac1n+O(n^{-2})\right)=\frac{\log n}{\log\alpha}+O(1),$$ so that $S_m=m\log m/\log\alpha+O(m)$ as $m\to\infty$, and the same holds for $K_m$.
