Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that  $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$? 

Iif $a(n)=\sqrt{ 2\sqrt {3\sqrt {4...\sqrt{n}}}}$‎,
‎$$a(n)^{2}= 2\sqrt {3\sqrt {4...\sqrt{n}}} \tag{1}$$‎
‎$$\dfrac{1}{2^{2}}.a(n)^{2}= 3\sqrt {4...\sqrt{n}} \tag{2}$$‎
‎$$\dfrac{1}{3^{2}}.\dfrac{1}{2^{2^{2}}}.a(n)^{2^{2}}= 4\sqrt{...\sqrt{n}} \tag{3}$$‎
$$\vdots$$ 
‎$$\prod_{i=0}^{n-2}\Big(\dfrac{1}{(n-i)^{2^{i}}}\Big).a(n)^{2^{n-1}}=1 \tag{n}$$‎
Then $$a(n)=(\prod _{i=0}^{n-2}{(n-i)^{2^{i}})}^{\dfrac{1}{2^{n-1}}}   \tag{1}$$‎
‎and‎
‎$$a(n+1)=(\prod _{i=0}^{n-1}{(n+1-i)^{2^{i}})}^{\dfrac{1}{2^{n}}} \tag{2}$$‎
From $(1)$ and $2$‎, 
‎$$\frac{a(n+1)}{a(n)}=(n+1)^{\frac{1}{2^{n}}}$$‎

‎Mathematica could not find the limit sequence for $n\longrightarrow \infty$‎.
