$$\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots\sqrt{(n-1)\sqrt{n}}}}}}$$
I know how to deal with roots like that when there is addition (like this: How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists?), but I have no idea what to do with multiplying.
The wanted quantity is not three. It is $$ \exp\sum_{n\geq 1}\frac{\log(n+1)}{2^n} $$ and the inner series can be easily approximated via summation by parts and creative telescoping:
$$ \begin{eqnarray*}S&=&\lim_{N\to +\infty}\sum_{n=1}^N\frac{\log(n+1)}{2^n}\\&=&\lim_{N\to +\infty}\left[(1-2^{-N})\log(N+1)-\sum_{n=1}^{N-1}\left(1-2^{-n}\right)\log\left(1+\tfrac{1}{n+1}\right)\right]\\&=&\log(2)+\sum_{n\geq 1}\frac{1}{2^n}\log\left(1+\tfrac{1}{n+1}\right)=\log(2)+2\sum_{n\geq 2}\frac{\log\left(1+\tfrac{1}{n}\right)}{2^n}\end{eqnarray*} $$ approximately equals $$\begin{eqnarray*}&& \log(2)+2\sum_{n\geq 2}\frac{1}{2^n}\left[\frac{1}{n+\tfrac{1}{2}}+\frac{1}{12\left(n-\tfrac{1}{2}\right)\left(n+\tfrac{1}{2}\right)\left(n+\tfrac{3}{2}\right)}\right]\\&=&\log(2)-\frac{143}{30}+\frac{49}{6\sqrt{2}}\log(1+\sqrt{2})\end{eqnarray*}$$ which is a number slightly larger than $1$. It follows that $$ C=\sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{\ldots}}}}} $$ is slightly larger than $e$, but definitely less than $3$. By Frullani's theorem we also have the following integral representation for $C$:
$$ C = \exp \int_{0}^{+\infty}\frac{2(1-e^{-x})}{x(2e^x-1)}\,dx\leq \exp\int_{0}^{+\infty} 2 e^{-5x/2}\left(1+\tfrac{25}{24}x^2\right)\,dx = \exp\left(\tfrac{16}{15}\right). $$
Actually products are easier because their logarithms are sums. In this case the result is $\exp\sum_{n\ge 2}2^{-n}\ln n$. The polylogarithm $\operatorname{Li}_s(z):=\sum_{n\ge 1}z^n n^{-s}$ satisfies $-\partial_s\operatorname{Li}_s(z):=\sum_{n\ge 2}z^n n^{-s}\ln n$, so your expression is $\exp-\partial_s\operatorname{Li}_s(z)|_{s=0,\,z=1/2}$.
n\to\infty. – J.G. Nov 11 '18 at 17:25