Irrationality of √2√3√4⋯\sqrt{2\sqrt{3\sqrt{4\cdots}}}

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.