Proving: $\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$ I need help to prove the following inequality with a nested radical.
$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$$
I have tired to use the Stirling approximation, but I got stuck with the nested radical.
Does anyone have any idea?
 A: We have that
$$\sqrt{1!\sqrt{2!\sqrt{\cdots\sqrt{\text{n}!}}}}=\prod_{k=1}^n(k!)^{1/2^k}=\prod_{1\le j\le k\le n}j^{1/2^k}=\prod_{j=1}^nj^{\sum_{k=j}^n1/2^k}=\prod_{j=1}^nj^{(1/2)^{j-1}-(1/2)^n}\\=\exp\left(\sum_{j=1}^n\ln j\left(\frac1{2^{j-1}}-\frac1{2^n}\right)\right)>\exp\left(\left(\sum_{j=2}^n\left(\frac12\right)^{j-1}\right)-\frac{n-1}{2^n}\right)\to\exp(1-0)=e$$
A: Following up on Peter's comment below the OP, we have
$$\begin{align}
\log\left(\sqrt{1!\sqrt{2!\sqrt{\ldots\sqrt{n!}}}}\right)
&={1\over2}\log(1!)+{1\over2^2}\log(2!)+\cdots+{1\over2^n}\log(n!)\\
&=\left({1\over2}+{1\over4}+\cdots+{1\over2^n}\right)\log1+\left({1\over4}+{1\over8}+\cdots+{1\over2^n}\right)\log2+\cdots+{1\over2^n}\log n\\
&\to\log1+{1\over2}\log2+{1\over4}\log3+{1\over8}\log4+\cdots\\
&\gt{1\over2}\log2+{1\over4}\log2+{1\over8}\log2+\cdots=\log2
\end{align}$$
So the requested inequality is not true.
A: You might consider this.
$$\sqrt{1!\sqrt{2!\sqrt{\cdots\sqrt{\text{n}!}}}}=\left(1!\right)^\frac{1}{2}\times\left(2!\right)^\frac{1}{4}\times\cdots\times\left(\text{n}!\right)^\frac{1}{2^\text{n}}=\exp\left\{\sum_{\text{k}=1}^\text{n}\frac{1}{2^\text{k}}\cdot\ln\left(\text{k}!\right)\right\}\tag1$$
Can you bound the sum inside.  It is convergent when $\text{n}\to\infty$?

$$\lim_{\text{k}\to\infty}\left|\frac{\frac{1}{2^{\text{k}+1}}\cdot\ln\left(\left(\text{k}+1\right)!\right)}{\frac{1}{2^\text{k}}\cdot\ln\left(\text{k}!\right)}\right|=\frac{1}{2}<1\tag2$$
