I can show the convergence of the following infinite product and some bounds for it:
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots<$$
$$<\left(1+\frac{1}{4} \right)\left(1+\frac{1}{9} \right)\left(1+\frac{1}{16} \right)\cdots=\prod_{k \geq 2} \left(1+\frac{1}{k^2} \right)=\frac{\sinh \pi}{2 \pi}=1.83804$$
Here I used Euler's product for $\frac{\sin x}{x}$.
The next upper bound is not as easy to evaluate, but still possible, taking two more terms in Taylor's series for $\sqrt[k]{1+\frac{1}{k} }$:
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{k-1}{2k^4}+\frac{2k^2-3k+1}{6k^6} \right)=$$
$$=\prod_{k \geq 2} \left(1+\frac{1}{k^2}-\frac{1}{2k^3}+\frac{5}{6k^4}-\frac{1}{2k^5}+\frac{1}{6k^6} \right)<$$
$$<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{108}+\frac{\pi^6}{5670}-1-\frac{\zeta (3)}{2}-\frac{\zeta (5)}{2} \right)=1.81654$$
The numerical value of the infinite product is approximately:
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=1.758743628$$
The ISC found no closed from for this number.
Is there some way to evaluate this product or find better bounds in closed form?
Edit
Clement C suggested taking logarithm and it was a very useful suggestion, since I get the series:
$$\ln \prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}= \frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots$$
I don't know how to find the closed form, but I can certainly use it to find the boundaries (since the series for logarithm are very simple).
$$\frac{1}{2} \ln \left(1+\frac{1}{2} \right)+\frac{1}{3} \ln \left(1+\frac{1}{3} \right)+\dots>\sum^{\infty}_{k=2} \frac{1}{k^2}-\frac{1}{2}\sum^{\infty}_{k=2} \frac{1}{k^3}$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}-\frac{1}{2}-\frac{\zeta (3)}{2} \right)=1.72272$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{5}{6}-\frac{\zeta (3)}{2} \right)=1.77065$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}-\frac{7}{12}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.75438$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}<\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{47}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4}\right)=1.76048$$
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}>\exp \left( \frac{\pi^2}{6}+\frac{\pi^4}{270}+\frac{\pi^6}{4725}-\frac{37}{60}-\frac{\zeta (3)}{2} -\frac{\zeta (5)}{4} -\frac{\zeta (7)}{6}\right)=1.75803$$
This method generates much better bounds than my first idea. The last two are very good approximations.
Edit 2
Actually, would it be correct to write (it gives the correct value of the product):
$$\prod_{k \geq 2}\sqrt[k]{1+\frac{1}{k}}=\frac{1}{2} \exp \left( \sum_{k \geq 2} \frac{(-1)^k \zeta(k)}{k-1} \right)$$
Best Answer
This is not an answer, but it's important and I post it separately from the question itself.
I found in this answer by @RandomVariable the following series:
$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\frac{\pi^2}{4}-1-4\int_{0}^{\pi/2} \frac{t~dt}{e^{\pi \tan t}+1} $$
They are related to $\gamma_1$ - Stieltjes constant.
This same series also appeared in this paper by Steven Finch, page 5.
$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=1.2577468869$$
This is the same numerical value as:
$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=1.2577468869$$
Which is confirmed in this paper by the same author, page 3, where this form of the series is used.
It is connected to the integral (page 2, the same paper):
$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=-\int_{1}^{\infty} \frac{\ln (y-[y])}{y^2}dy$$
Where $[y]$ is the floor function, meaning $y-[y]$ is the fractional part of $y$.
In another paper this series is connected to the numer of divisors of $n!$, however slightly different integral representation is used (page 3):
$$\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\int_{1}^{\infty} \frac{\ln ([y]+1)}{y^2}dy$$
And finally, this is slightly related to Alladi-Grinstead Constant, which is given by:
$$e^{c-1}$$
$$c=\sum_{k=2}^{\infty} \frac{\ln (\frac{k}{k-1})}{k}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k+1}=0.788530566$$
See also the original Alladi and Grinstead paper here.
And this is also somehow connected to the Luroth series representations of real numbers.
Oh, and thanks to @SteveKass for this useful link.
Comparing the convergence of three series, we find that even though they are equivalent, the convergence rate is drastically different.
$$\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}$$
We can also obtain the following interesting equality:
$$(1+1)\sqrt{1+\frac{1}{2}} \sqrt[3]{1+\frac{1}{3}} \sqrt[4]{1+\frac{1}{4}} \cdots=\sqrt{2} \sqrt[6]{3} \sqrt[12]{4} \sqrt[20]{5} \sqrt[30]{6} \cdots=\prod_{k=1}^{\infty}(k+1)^{\frac{1}{k(k+1)}}$$