Is $\sqrt {2 \sqrt {3 \sqrt {4 \ldots}}}$ algebraic or transcendental?
Solution 1:
This is not an answer, just a list of papers concerned with this constant; it's too long for a comment. Going by the reviews of the papers, they deal with efficient methods for calculating the constant, and not (directly) with questions of irrationality and/or transcendence. The first paper on the list is too new to have been reviewed at this point.
MR3349435
Lu, Dawei; Song, Zexi;
Some new continued fraction estimates of the Somos' quadratic recurrence constant.
J. Number Theory 155 (2015), 36–45.
MR3019753
Chen, Chao-Ping
New asymptotic expansions related to Somos' quadratic recurrence constant.
C. R. Math. Acad. Sci. Paris 351 (2013), no. 1-2, 9–12.
MR2825112 (2012h:11181)
Hirschhorn, Michael D.
A note on Somos' quadratic recurrence constant.
J. Number Theory 131 (2011), no. 11, 2061–2063.
MR2809034 (2012e:05038)
Nemes, Gergő
On the coefficients of an asymptotic expansion related to Somos' quadratic recurrence constant.
Appl. Anal. Discrete Math. 5 (2011), no. 1, 60–66.
MR2684487 (2011i:11179)
Mortici, Cristinel
Estimating the Somos' quadratic recurrence constant.
J. Number Theory 130 (2010), no. 12, 2650–2657.
MR2319662 (2008f:40013)
Sondow, Jonathan; Hadjicostas, Petros
The generalized-Euler-constant function $\gamma(z)$ and a generalization of Somos's quadratic recurrence constant.
J. Math. Anal. Appl. 332 (2007), no. 1, 292–314.
MR2262724 (2008b:11081)
Hessami Pilehrood, Khodabakhsh; Hessami Pilehrood, Tatiana
Arithmetical properties of some series with logarithmic coefficients.
Math. Z. 255 (2007), no. 1, 117–131.
Solution 2:
There exists a finite difference formulation of this problem. This doesn't solve your problem, but it may be at least a bit enlightening, Consider $f(x)$ such that
$$ f(x) = x\sqrt{f(x+1)} $$
Then it follows
$$ f(x) = x \sqrt{(x+1) \sqrt{(x+2) \sqrt{... }}} $$
And $f(1)$ is the value you wish to evaluate.
Another way to write that is
$$ \frac{f(x)^2}{x^2} = f(x+1) $$
Let $f(x) = 2^{g(x)} $
$$ \frac{2^{2g(x)}}{x^2} = 2^{g(x+1} $$
Taking the logarithm of both sides base-2
$$ 2g(x) - 2\log_2{x} = g(x+1) $$
We can now evaluate, this as a run of the mill finite difference equation in G
$$ D_{1,x}g - g = -2\log_2{x} $$
(... lots of work that you can comment and request if you want ...)
$$ 2^{1-x}g = -D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right] $$
So it follows that
$$ g = -2^{x-1}D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right] $$
And that
$$ f = 2^{ -2^{x-1}D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right]} $$
Note that $f(0) = 0$ and therefore the expression
$$ D_{1,x}^{-1}\left[2^{1-x}\log_2{x} \right] $$
must tend to positive infinity as $x$ tends to 0. Analytically i'm not sure how to continue this. But putting it into Wolfram Alpha yields:
http://www.wolframalpha.com/input/?i=f%28x%2B1%29+-+f%28x%29+%3D+2%5E%7B1-x%7Dlog_2%28x%29
That the expression is dependent on polylogarithms (unsurprising) and something called the lerch transcendent (I believe these have to do with generalizing the riemann zeta function).
So if you choose to go down the finite difference path, it seems that you might be using some heavy analytic number theory tools to determine if the expression is transcendental.