Inverse-square relationship and seeing stars from far away: Is there maximum distance from which we can see a star? The question is:
Taking into account as obstacle only inverse-square relationship. Does there exist border distance below which we can see a star and above which we can not?

From my understanding in some great (infinite?) distance from radiation emitter, there would be single photon for few square kilometers or even millions of square kilometers (depending on distance). If I am right, we would not be able to see object emitting it. Am I right?
If so, are there potential objects in the cosmos we are not seeing because of the law and not because their light did not reach us yet?
How far would these objects have to be (taking some average star as an example)?
Edit:
Answering these questions, take into consideration technology which is available today please. :)
 A: You can look here to see the most distant astronomical objects ever seen. Obviously, it depends on the luminosity of the object. The brighter it is, the farther away it can be seen from; that's why all the farthest objects are galaxies. They are over 13 Gly from earth (~$9.5\dot{}10^{24}$ m).
The farthest star we've ever seen is apparently SDSS (Sloan Digital Sky Survey) J1229+1122, which is 55 Mly from earth (~$5.2\dot{}10^{23}$ m). However, that is a blue supergiant, not an average star.
I don't think this answers your question though. Let's look at how far away an average star would need to be to have the same brightness as the faintest thing we've ever observed. The faintest object we've ever recorded has an apparent magnitude of ~+31.5 $^{[1]}$ (the magnitude scale is weird, it goes backwards and is logarithmic).
If we take the Sun, a very average star, how far away would it need to be before its apparent magnitude "dropped" to +31.5? The distance modulus equation is
$$
m-M=5 \log_{10} (\frac{r}{10})
$$
where $m$ is the apparent magnitude, $M$ is the absolute magnitude, and $r$ is the distance in parsecs. The sun has an absolute magnitude $M_\odot = 4.83$, and so if we set $m_\odot=31.5$, 
$$
26.67=5 \log_{10} (\frac{r}{10})
$$
$$
r=2158000 \textrm{ pc}
$$
So, the sun would need to be $2158000$ pc ($6.7\dot{}10^{22}$ m) away from us to have the same magnitude as the faintest object ever seen.

1 https://arxiv.org/abs/1305.1931, from Wikipedia
