<s>[INST] How to prove :$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$ In  fact initially I wanted to prove that
$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <2$$
Which by the accepted answer here fails to be true.

@Barry Cipra advised me to ask this question in a different post:
How can I now prove that:
$$\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} <3$$

Note that the answers here do not provide an estimate of this sequence
Can anyone have any idea?
 [/INST] For $n\geq 2$,
\begin{align*}
\sqrt{1!\sqrt{2!\sqrt{3!\sqrt{\cdots\sqrt{n!}}}}} & = 2^{1/4+1/8+1/16+\ldots+1/2^n} \cdot 3^{1/8+1/16+1/32+\ldots+1/2^n}\cdot \ldots \cdot n^{1/2^n} \\
& < 2^{1/2}\cdot 3^{1/4} \cdot 4^{1/8} \cdot \ldots \cdot n^{1/2^{n-1}} \\
& = \sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}} \\
&<3\end{align*}
The last inequality was proved here.
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