Irrationality of $\sqrt{2\sqrt{3\sqrt{4\cdots}}}$ In this question it is stated that Somos' quadratic recurrence constant
$$\alpha=\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots}}}}$$
is an irrational number. [update: the author of that question is no longer claiming to have a proof of this]
This fact seems by no means trivial to me. The algebraic numbers $\sqrt{2}$, $\sqrt{2\sqrt{3}}$, $\sqrt{2\sqrt{3\sqrt{4}}}$, $\dots$ do not converge quickly enough to $\alpha$, so one cannot reuse the proof of Liouville's theorem in this case.
Approximation arguments do not seem a good way, since
$$ \sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots}}}}=2$$
is rational instead!
What am I missing?
 A: Your original post implies you have thought of, or know of, a good way of using Liouville's theorem to solve the problem. Here's the start of some ideas on the issue that perhaps you will be able to finish off.
We can write $\alpha$ in the following way:
$$\alpha=\prod_{n=1}^\infty\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}.$$
We will also define the partial product
$$\alpha_N=\prod_{n=1}^N\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}.$$
Then, for some integers $p$, $q$:
$$\left|\alpha-\frac{p}{q}\right|\leq\left|\alpha-\alpha_N\right|+\left|\alpha_N-\frac{p}{q}\right|$$.
Now, the first of these values can be quickly bounded:
\begin{eqnarray}\alpha-\alpha_N&=&\alpha_N\left(\prod_{n=N+1}^\infty\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}-1\right)\\
&\leq& \alpha_N\left(\prod_{n=N+1}^\infty\left(2\right)^{2^{-(n-1)}}-1\right)\\
&\leq& \alpha_N\left(2^{2^{-(N-1)}}-1\right) \\ 
&\leq& 4\alpha_N\cdot 2^{-N} \\
&\leq& 4\alpha\cdot 2^{-N}
\end{eqnarray}
However, I can't see how to bound the other expression, $\left|\alpha_N-\frac{p}{q}\right|$. Because the $\alpha_N$ are algebraic, we can't instantly point to the existence of some rationals that approximate them ``well enough'' for our purposes here, as far as I can see.
A: I may be completely out of line here and my thinking doesn't seem to be at the level of most of the other proofs presented, but I wonder if the following could lead to a proof. Please do not consider this a rigorous proof by any stretch.
First, use the distributive property of roots to separate $\sqrt{2}$ from $\sqrt{\sqrt{3\sqrt{4\sqrt{...}}}}$ which gives us $\sqrt{2}*\sqrt{\sqrt{3\sqrt{4\sqrt{...}}}}$ Now, most have seen the proof from contradiction that $\sqrt{2}$ is irrational. Would it not then just be an issue of showing that no matter the remaining sequence, when multiplied by $\sqrt{2}$, it can never produce a rational number? Essentially, an irrational number cannot be multiplied by any such number as to produce a rational number $\frac{p}{q}$. Am I completely off base here?
